Chapter 3: Motion in a Plane
For centuries, gunners across Europe fired cannonballs by trial and error. They knew that tilting the barrel higher sent the ball further — up to a point — but nobody could say what curve the ball actually followed. Many believed it travelled in a straight slanting line, then turned and dropped almost vertically to the ground.
In the early 1600s, in Italy, Galileo Galilei looked at the problem in a completely new way. Instead of treating the flight as one complicated motion, he split it into two simple ones happening at the same time. Sideways, he argued, nothing pushes or pulls the ball, so it keeps moving with a steady horizontal speed. Downwards, gravity pulls it continuously, so its vertical motion speeds up exactly like that of a falling stone. The ball is doing both at once, and neither motion interferes with the other.
When Galileo combined these two independent motions mathematically, a familiar shape appeared: a parabola. The curve of a cannonball, a cricket ball hit for a sixer, and a jet of water from a hose are all the same curve. This single idea — that motion in a plane can be broken into two independent motions along perpendicular directions — is the key that unlocks this entire chapter.
Diagram
Fig. 3.1 – A ball launched at an angle tracing a parabolic path, with its constant horizontal velocity and its changing vertical velocity shown as separate arrows at three points along the curve.
By the end of this chapter, you will be able to answer each of these questions using the physics you learn here.
3.1 Introduction
Look around you and you will find objects in motion everywhere — a fan blade turning, a bus pulling away from a stop, a leaf drifting down from a tree. In physics, we describe all of this with one simple idea: motion is a change in the position of an object with time. If an object’s position today is the same as it was a moment ago, it is at rest; if the position keeps changing, it is in motion.
Notice that this definition contains two things — position and time. To describe motion completely, you must always answer both questions: where is the object, and when is it there? Every quantity you will meet in this chapter (displacement, velocity, acceleration) is built out of these two ideas.
Three Kinds of Motion
Think of a toy car being pushed along a straight line, a cricket ball hit by a batsman for a sixer, and an aeroplane flying from one city to another. These three familiar examples are not just different examples of motion — they belong to three different categories of motion.
The motion of objects can be divided into three categories:
- Motion along a straight line, that is, rectilinear motion — the toy car pushed in one particular direction.
- Motion in two dimensions, that is, motion in a plane — the cricket ball, which rises, moves forward and comes down, all within a single vertical plane.
- Motion in three dimensions, that is, motion in space — the aeroplane, which can move forward, sideways and up or down.
Diagram
Fig. 3.2 – Three side-by-side sketches: a toy car on a straight line, a cricket ball tracing a curve in a vertical plane, and an aeroplane changing height and direction in space.
The number of dimensions simply tells you how many independent directions you need in order to state where the object is. On a straight line, one number (say the value of \(x\)) is enough. In a plane, you need two numbers (\(x\) and \(y\)). In space, you need three (\(x\), \(y\) and \(z\)). This is why we speak of one-, two- and three-dimensional motion.
What You Already Know About Rectilinear Motion
You have studied rectilinear motion in earlier standards. Its special feature is that the force acting on the object and the velocity of the object are both along one and the same line. Because everything happens along a single line, direction becomes very easy to handle: the distances are measured along that line only, and we can indicate distances along the +ve and –ve axes as being positive and negative respectively.
In other words, in rectilinear motion a plus sign or a minus sign is enough to carry all the information about direction. This is a convenience you will lose the moment the motion leaves the straight line, because in a plane a direction can no longer be described by just a sign — it needs a full vector.
Why Vectors Make Life Easier
The study of the motion of an object in a plane or in space becomes much easier, and the corresponding equations become more elegant, if we use vector quantities. A vector carries magnitude and direction together in a single symbol, so one vector equation can replace two or three separate equations written for the separate directions.
This is why, in this chapter, we will first recall the basic facts about rectilinear motion, but we will now write them using vector notation. You already know these ideas; what is new is the language. Practising this language on the simple case of straight-line motion will make it far easier when we move on to motion in two dimensions.
The Plan of This Chapter
After recalling rectilinear motion in vector form, we will then study motion in two dimensions. Within two-dimensional motion, our detailed treatment will be restricted to projectile motion only — the motion of an object thrown into the air and moving under gravity.
Finally, circular motion, that is, the motion of an object around a circular path, will be introduced here. You will meet the basic ideas of uniform circular motion in this chapter, and it will be studied in detail in the next standard.
3.2 Rectilinear Motion
Consider an object moving along a straight line. Let us assume this line to be along the \(x\)-axis. Let \(\vec{x}_1\) and \(\vec{x}_2\) be the position vectors of the body at times \(t_1\) and \(t_2\) during its motion.
A position vector simply tells us where the object is, measured from a chosen origin. Because the motion here is along a single line, the position vector points either along the \(+x\) direction or along the \(-x\) direction, and its magnitude is the distance of the object from the origin. Once we know the position at two different times, we can define every other quantity of motion.
The following quantities can be defined for the motion.
Displacement
The displacement of the object between \(t_1\) and \(t_2\) is the difference between the position vectors of the object at the two instances. Thus, the displacement is given by
\[\vec{s} = \Delta\vec{x} = \vec{x}_2 - \vec{x}_1 \qquad \text{--- (3.1)}\]
Here \(\vec{x}_1\) is the initial position vector, \(\vec{x}_2\) is the final position vector, and \(\vec{s}\) is the displacement. Its SI unit is the metre (m). Its direction is along the line of motion of the object. Its dimensions are that of length, i.e. \([\mathrm{L^1 M^0 T^0}]\).
Notice what this definition does not contain: it says nothing about the route taken. Displacement cares only about where you started and where you finished.
For example, if an object has travelled through 1 m from time \(t_1\) to \(t_2\) along the \(+ve\) \(x\)-direction, the magnitude of its displacement is 1 m and its direction is along the \(+ve\) \(x\)-axis. On the other hand, if the object travelled along the \(+ve\) \(y\) direction through the same distance in the same time, the magnitude of its displacement is the same as before, i.e. 1 m, but the direction of the displacement is along the \(+ve\) \(y\)-axis.
The two displacements have equal magnitudes but different directions, and therefore they are different displacements. This is exactly why displacement must be treated as a vector and not as a plain number.
Path Length
Path length is the actual distance travelled by the object during its motion. It is a scalar quantity and its dimensions are also that of length, \([\mathrm{L^1 M^0 T^0}]\), with the SI unit metre (m).
Unlike displacement, path length keeps a record of every metre the object actually travels, including any back-and-forth movement. Let us follow one object step by step and watch the two quantities drift apart.
If an object travels along the \(x\)-axis from \(x = 2\) m to \(x = 5\) m, then the distance travelled is 3 m. In this case the displacement is also 3 m and its direction is along the \(+ve\) \(x\)-axis. So far, the two agree — because the object moved in one direction only.
However, if the object now comes back to \(x = 4\), then the distance through which the object has moved increases to \(3 + 1 = 4\) m. Its initial position was \(x = 2\) m and the final position is now \(x = 4\) m, and thus its displacement is \(\Delta x = 4 - 2 = 2\) m, i.e. the magnitude of the displacement is 2 m and its direction is along the \(+ve\) \(x\)-axis. The moment the object reversed, path length kept adding while displacement shrank.
If the object now moves to \(x = 1\), then the distance travelled, i.e. the path length, increases to \(4 + 3 = 7\) m, while the magnitude of displacement becomes \(2 - 1 = 1\) m and its direction is along the negative \(x\)-axis.
Diagram
Fig. 3.3 – A number line marked from x = 1 m to x = 5 m showing the object’s journey 2 → 5 → 4 → 1 with curved arrows above, and the final displacement shown as a single straight arrow from 2 to 1 below the line.
Average Velocity
Average velocity is defined as the displacement of the object during the time interval over which average velocity is being calculated, divided by that time interval. As displacement is a vector quantity, the velocity is also a vector quantity. Its dimensions are \([\mathrm{L^1 M^0 T^{-1}}]\) and its SI unit is m/s.
If the position vectors of the object are \(\vec{x}_1\) and \(\vec{x}_2\) at times \(t_1\) and \(t_2\) respectively, then the average velocity is given by
\[\vec{v}_{av} = \frac{\vec{x}_2 - \vec{x}_1}{(t_2 - t_1)} \qquad \text{--- (3.2)}\]
The physical meaning is worth stating plainly: average velocity tells you the net rate at which the object shifted its position, ignoring completely what it did in between. A dividing by a scalar (time) never changes direction, so \(\vec{v}_{av}\) points in exactly the same direction as the displacement \(\Delta \vec{x}\).
For example, if the positions of an object are \(x = +2\) m and \(x = +4\) m at times \(t = 0\) and \(t = 1\) minute respectively, the magnitude of its average velocity during that time is \(v_{av} = (4 - 2)/(1 - 0) = 2\) m per minute, and its direction will be along the \(+ve\) \(x\)-axis. We write
\[\vec{v}_{av} = 2\hat{i} \ \text{m/min}\]
where \(\hat{i}\) is a unit vector along the \(x\)-axis. A unit vector has magnitude 1 and no units of its own; its only job is to carry the direction, leaving the number in front to carry the magnitude.
Average Speed
Average speed is defined as the total path length travelled during the time interval over which average speed is being calculated, divided by that time interval.
\[\text{Average speed} = v_{av} = \frac{\text{path length}}{\text{time interval}}\]
It is a scalar quantity and has the same dimensions as that of velocity, i.e. \([\mathrm{L^1 M^0 T^{-1}}]\), with SI unit m/s.
If the rectilinear motion of the object is only in one direction along a line, then the magnitude of its displacement will be equal to the distance travelled, and so the magnitude of average velocity will be equal to the average speed.
However, if the object reverses its direction (the motion remaining along the same line), then the magnitude of displacement will be smaller than the path length, and the average speed will be larger than the magnitude of average velocity.
Instantaneous Velocity
Average velocity smooths out everything that happens inside the time interval. But often we want to know how fast the object is moving right now — at one particular instant. That is instantaneous velocity.
Instantaneous velocity of an object is its velocity at a given instant of time. It is defined as the limiting value of the average velocity of the object over a small time interval (\(\Delta t\)) around \(t\) when the value of the time interval (\(\Delta t\)) goes to zero.
\[\vec{v} = \lim_{\Delta t \to 0}\left(\frac{\Delta \vec{x}}{\Delta t}\right) = \frac{d\vec{x}}{dt} \qquad \text{--- (3.3)}\]
\(\dfrac{d\vec{x}}{dt}\) is the derivative of \(\vec{x}\) with respect to \(t\) (see Chapter 2). Its SI unit is m/s.
The idea behind the limit is straightforward. Take a shorter and shorter time interval around the instant you care about. As the interval shrinks, the object has less and less time to change its velocity inside it, so the average over that interval becomes a better and better description of the single instant. The limit is the exact value.
Instantaneous Speed
Instantaneous speed is the speed of an object at a given instant of time \(t\). It is the limiting value of the average speed of the object taken over a small time interval (\(\Delta t\)) around \(t\) when the time interval goes to zero.
In such a limit, the path length will be equal to the magnitude of the displacement, and so the instantaneous speed will always be equal to the magnitude of the instantaneous velocity of the object.
Graphical Study of Motion
We can study the motion of an object by using graphs showing its position as a function of time. Such a graph is called an \(x\)-\(t\) graph, and its great advantage is that the slope of the graph at any point is the velocity at that moment — because velocity is precisely the rate of change of position with time. Figure 3.4 shows the graphs of position as a function of time for five different types of motion of an object.
Figure 3.4(a) shows an object at rest, for which the \(x\)-\(t\) graph is a horizontal straight line. Since the position is not changing, displacement of the object is zero. Velocity is displacement (which is zero) divided by time interval, or the derivative of displacement with respect to time. It can be obtained from the slope of the line plotted in the figure, which is zero.
Diagram
Fig. 3.4(a) – Position-time graph: a horizontal straight line, labelled “object at rest”.
Figure 3.4(b) shows the \(x\)-\(t\) graph for an object moving with constant velocity along the \(+ve\) \(x\)-axis. Since velocity is constant, displacement is proportional to elapsed time. The slope of the straight line is \(+ve\), showing that the velocity is along the \(+ve\) \(x\)-axis. As the motion is uniform, the average velocity is the same as the instantaneous velocity at all times. Also, the speed is equal to the magnitude of the velocity.
Diagram
Fig. 3.4(b) – Position-time graph: a straight line of positive slope rising from the origin, labelled “uniform velocity along +ve x-axis”.
Figure 3.4(c) shows the \(x\)-\(t\) graph for a body moving with uniform velocity but along the \(-ve\) \(x\)-axis, the slope of the line being \(-ve\).
Diagram
Fig. 3.4(c) – Position-time graph: a straight line of negative slope, falling with time, labelled “uniform velocity along -ve x-axis”.
Figure 3.4(d) shows the \(x\)-\(t\) graph of an object having oscillatory motion with constant speed. The direction of velocity changes from \(+ve\) to \(-ve\) and vice versa over fixed intervals of time. This is why the graph is a zig-zag: each straight segment has a constant slope, but the sign of the slope flips every time the object turns around.
Diagram
Fig. 3.4(d) – Position-time graph: a zig-zag (triangular wave) pattern of straight segments with alternately positive and negative slopes, labelled “oscillatory motion”.
Figure 3.4(e) shows the motion of an object with nonuniform velocity. Its velocity changes with time and, therefore, the average and instantaneous velocities are different. The figure shows the average velocity over the time interval from \(t_1\) to \(t_4\) around time \(t_0\), which can be seen from Eq. (3.2) to be the slope of the line AB. This line AB is a chord — a straight line joining two points on the curve — and the slope of a chord is always the average velocity between those two instants.
Diagram
Fig. 3.4(e) – Position-time graph: a smooth curve with chord AB drawn between t₁ and t₄, a shorter chord CD between t₂ and t₃, and the tangent PQ drawn at t₀.
For a smaller time interval from \(t_2\) to \(t_3\), the average velocity is the slope of the line CD. If we keep reducing the time interval around \(t_0\), we will ultimately come to a limit when the time interval will go to zero, and lines AB, CD… will go over to the tangent to the curve at \(t_0\). The instantaneous velocity at \(t_0\) will thus be equal to the slope of the tangent PQ at \(t_0\) (see Eq. (3.3)).
Notice that this graph is the picture of the limiting process described in Eq. (3.3). Each shorter chord hugs the curve more closely, and in the limit the chord becomes the tangent. The tangent is the graphical meaning of the derivative.
Acceleration
Acceleration is defined as the rate of change of velocity with time. It is a vector quantity and its dimensions are \([\mathrm{L^1 M^0 T^{-2}}]\), with SI unit m/s².
The average acceleration of an object having velocities \(\vec{v}_1\) and \(\vec{v}_2\) at times \(t_1\) and \(t_2\) is given by
\[\vec{a} = \frac{\left(\vec{v}_2 - \vec{v}_1\right)}{\left(t_2 - t_1\right)} \qquad \text{--- (3.4)}\]
Because velocity is a vector, “a change in velocity” can mean a change in its magnitude, a change in its direction, or both. In rectilinear motion only the magnitude can change, so acceleration is simply speeding up or slowing down. Later in this chapter, in circular motion, you will meet an acceleration that changes only the direction of the velocity and not its magnitude.
Instantaneous acceleration is the limiting value of the average acceleration when the time interval goes to zero. It is given by
\[\vec{a} = \lim_{\Delta t \to 0}\left(\frac{\Delta \vec{v}}{\Delta t}\right) = \frac{d\vec{v}}{dt} \qquad \text{--- (3.5)}\]
The instantaneous acceleration at a given time is the slope of the tangent to the velocity versus time curve at that time. This is the same graphical rule as before, applied one level up: slope of the \(x\)-\(t\) graph gives velocity, and slope of the \(v\)-\(t\) graph gives acceleration.
Figure 3.5 shows the velocity versus time (\(v\)-\(t\)) graphs for four different cases.
Figure 3.5(a) represents the motion of an object with zero acceleration, i.e. constant velocity. The shaded area under the velocity-time graph over some time interval \(t_1\) to \(t_2\), shown in Fig. 3.5(a), is equal to \(v_0 (t_2 - t_1)\), which is the magnitude of the displacement of the object from \(t_1\) to \(t_2\).
Diagram
Fig. 3.5(a) – Velocity-time graph: a horizontal line at v₀, with the rectangular area between t₁ and t₂ shaded.
Figure 3.5(b) is the velocity-time graph for an object moving with constant \(+ve\) acceleration (magnitude of velocity uniformly increasing with time). The graph is a straight line sloping upward, and its constant slope is the constant acceleration.
Diagram
Fig. 3.5(b) – Velocity-time graph: a straight line of positive slope rising from v₁ to v₂, with the area between t₁ and t₂ shaded.
Figure 3.5(c) shows similar motion but the object has \(-ve\) acceleration, i.e. the acceleration is opposite to the direction of velocity, which therefore decreases uniformly with time.
Diagram
Fig. 3.5(c) – Velocity-time graph: a straight line of negative slope falling from v₁ to v₂, with the area between t₁ and t₂ shaded.
The area under both the curves between two instants of time is the displacement of the object during that time interval (as shown below).
Figure 3.5(d) shows the motion of an object having nonuniform acceleration. The average acceleration between \(t_1\) and \(t_2\) around \(t_0\), and the instantaneous acceleration at \(t_0\) for the object, are shown by the straight lines AB and CD respectively.
Diagram
Fig. 3.5(d) – Velocity-time graph: a curve of increasing steepness, with chord AB drawn between t₁ and t₂ and the tangent CD drawn at t₀.
The area under the velocity-time curves in Figs. 3.5(a) to (d) can be written using the definition of the integral given in Chapter 2 as
\[\text{Area} = \int_{t_1}^{t_2} v\, dt = \int_{t_1}^{t_2}\frac{dx}{dt}\, dt = \int_{t_1}^{t_2} dx = x(t_2) - x(t_1) \qquad \text{--- (3.6)}\]
\[= \text{displacement of the object from } t_1 \text{ to } t_2.\]
The reasoning behind Eq. (3.6) is worth reading slowly. Velocity is \(dx/dt\). So \(v\, dt\) is a tiny piece of displacement covered in a tiny time \(dt\). Adding up (integrating) all these tiny pieces between \(t_1\) and \(t_2\) must give the total change in position, which is exactly \(x(t_2) - x(t_1)\). Graphically, each strip of width \(dt\) and height \(v\) is one such tiny piece, so the total area is the total displacement.
Equations of Motion for Uniform Acceleration
We can graphically derive Newton’s equations of motion for an object moving with uniform acceleration. Consider an object having position \(x = 0\) at \(t = 0\). Let the velocity at \(t = 0\) be \(u\) and at time \(t\) be \(v\). The graphical representation of motion is shown in Fig. 3.6.
Diagram
Fig. 3.6 – Velocity-time graph for uniform acceleration: line AB rising from A (at u on the v-axis) to B (at v at time t), with rectangle OACD and triangle ABC marked below it.
Since the acceleration is uniform, the \(v\)-\(t\) graph is the straight line AB. The acceleration is given by the slope of the line AB. Thus,
\[\text{Acceleration, } a = \frac{v - u}{t - 0} = \frac{v - u}{t}\]
\[\therefore \ v = u + at \qquad \text{--- (3.7)}\]
This is the first equation of motion. Here \(u\) is the initial velocity (m/s), \(v\) is the velocity after time \(t\) (m/s), \(a\) is the uniform acceleration (m/s²) and \(t\) is the time (s).
As we know, the area under the curve in the velocity-time graph is the displacement of the object. Thus displacement \(s\) = area of the quadrilateral OABD = area of triangle ABC + area of rectangle OACD.
The rectangle OACD has height \(u\) and width \(t\), giving area \(ut\). The triangle ABC has base \(t\) and height \((v - u)\), giving area \(\tfrac{1}{2}(v-u)t\). Therefore
\[s = \frac{1}{2}\left(v - u\right)t + ut\]
Using Eq. (3.7), \(v - u = at\), so the triangle’s area becomes \(\tfrac{1}{2}at^2\), and we obtain
\[s = ut + \frac{1}{2}at^2 \qquad \text{--- (3.8)}\]
This is the second equation of motion.
As the acceleration is constant, the velocity is increasing linearly with time, and we can use the average velocity \(v_{av}\) to calculate the displacement using Eq. (3.7) as
\[s = v_{av}\, t = \left(\frac{v + u}{2}\right)t = \frac{\left(v + u\right)\left(v - u\right)}{2a}\]
In the last step we replaced \(t\) by \((v - u)/a\), which comes straight from Eq. (3.7). Multiplying out the numerator gives \(v^2 - u^2\), so
\[\therefore s = \left(v^2 - u^2\right)/\left(2a\right)\]
\[\therefore v^2 - u^2 = 2a\cdot s \qquad \text{--- (3.9)}\]
This is the third equation of motion.
Vector notation was not included here as the motion was rectilinear. In a straight line, a plus or minus sign is enough to record direction, so the equations can be written with ordinary numbers. When we come to motion in a plane, these same three equations will have to be rewritten with arrows on top.
Free Fall
The most common example of uniform rectilinear motion with uniform acceleration of an object in day-to-day life is a freely falling body. When a body starts with zero velocity at a certain height from the ground and falls under the influence of the gravity of the Earth, it is said to be in free fall.
The only other force that acts on it is that of the air resistance or friction. For displacements of a few metres, this force is too small and can be neglected. The acceleration of the body is the acceleration due to gravity, which is along the vertical direction and can be assumed to be constant over distances which are small compared to the radius of the Earth.
Thus the velocity and acceleration are both along the vertical direction, and the motion is a uniform rectilinear motion with uniform acceleration. Because both vectors lie along the same vertical line, all three equations of motion apply directly, with \(a\) replaced by \(g\).
Relative Velocity
You must have often experienced relative motion. The most striking example is when you are going in a train and another train travelling in the same direction along parallel tracks overtakes you. If you look at that train, it actually seems to be moving much slower than what your train seemed to move — and yet it is overtaking you.
On the other hand, if your train overtakes another train travelling on a parallel track in the same direction, and you look at that train, you feel that your train has suddenly slowed down. Why does this happen?
This is because when you look at the neighbouring train, you are actually experiencing relative motion, i.e. your motion with respect to the other train, or the motion of the other train with respect to you.
Thus, in the first case, as the other train overtakes you, what you perceive is the velocity of the train with respect to you, i.e. the difference in the velocities of the two trains, which most often is much smaller than the velocity of your train. In the second case, you are moving faster, but when you look at that train you only feel your velocity relative to it, and therefore your velocity appears to be lower than its actual value.
We can define the relative velocity of object A with respect to object B as the difference between their velocities, i.e.
\[v_{AB} = v_A - v_B \qquad \text{--- (3.10)}\]
Similarly, the velocity of B with respect to A is given by
\[v_{BA} = v_B - v_A \qquad \text{--- (3.11)}\]
Comparing the two equations, \(v_{AB} = -v_{BA}\): the two relative velocities have equal magnitudes but opposite directions. If A sees B moving east at 10 m/s, then B sees A moving west at 10 m/s.
We assume that at time \(t = 0\), A and B were at the same point \(x = 0\). As they are travelling with different velocities, the distance between them will go on increasing with time in direct proportion to the difference in their velocities, i.e. the relative velocity between them.
3.3 Motion in Two Dimensions - Motion in a Plane
So far we were considering rectilinear motion of an object. The direction of motion of the object was always along one straight line, and that made life easy: a plus or a minus sign was enough to describe direction.
Now we will consider the motion of an object in two dimensions, i.e. along a plane. Here, the direction of the force acting on an object will not be in the same line as its initial velocity. Thus, the velocity and acceleration will have different directions.
This single change is what makes plane motion genuinely new. Think of a ball thrown at an angle: its initial velocity points up and forward, while gravity pulls it straight down. The force and the velocity are no longer along the same line, so the object cannot simply speed up or slow down — it must also turn. A curved path is the result.
For this reason we have to use vector equations. A vector equation carries direction information built into it, so a single equation can handle motion that is happening along two directions at once.
Diagram
Fig. 3.7 – A ball moving along a curved path in the x-y plane, with its velocity vector drawn along the tangent at one point and the force (and hence acceleration) vector drawn in a clearly different direction at the same point.
What Carries Over, and What Changes
The definitions of various terms given in Section 3.2 will remain valid. Displacement is still the change in position vector; average velocity is still displacement divided by time; instantaneous velocity is still the derivative of the position vector; acceleration is still the rate of change of velocity. You do not have to learn a new set of definitions.
There is, however, one important difference. The magnitude of the average velocity and the value of the average speed will be different, as the magnitude of the displacement need not be equal to the path length.
In rectilinear motion, these two became different only when the object reversed its direction. In a plane, they differ whenever the path is curved — and a curved path is the normal case here, not the exception.
For example, if a particle travels along a circle and comes back to its original position, its displacement will be zero but the path length will be equal to the circumference of the circle. Its average velocity for the full round is therefore zero, while its average speed is definitely not.
3.3.1 Average and Instantaneous Velocities
For studying the motion of an object in two dimensions, for simplicity, we will take the plane to be the \(x\)-\(y\) plane. To describe the position of an object in this plane we will have to specify both its \(x\) and \(y\) coordinates — one number is no longer enough.
The definitions of displacement, average and instantaneous velocities, average and instantaneous speeds, and acceleration will be the same as those for rectilinear motion, except that each of these quantities will now have components along the \(x\) and \(y\) directions.
Position Vector and Displacement in a Plane
Let us assume the object to be at point P at time \(t_1\), as shown in Fig. 3.8(a).
Diagram
Fig. 3.8(a) – Motion in two dimensions: a curved path in the x-y plane with points P and Q marked, position vectors r₁ and r₂ drawn from the origin, the displacement vector PQ joining them, and the components Δx and Δy shown.
The position of the object will be described by its position vector \(\vec{r}_1\). This can be written in terms of its components along the \(x\) and \(y\) axes as
\[\vec{r}_1 = x_1\hat{i} + y_1\hat{j} \qquad \text{--- (3.12)}\]
Here \(x_1\) and \(y_1\) are the coordinates of P in metres, while \(\hat{i}\) and \(\hat{j}\) are unit vectors along the \(x\) and \(y\) axes respectively. Reading this equation in words: “to reach P from the origin, walk \(x_1\) metres along \(x\), then \(y_1\) metres along \(y\).”
At time \(t_2\), let the position of the object be Q and its position vector be \(\vec{r}_2\).
\[\vec{r}_2 = x_2\hat{i} + y_2\hat{j} \qquad \text{--- (3.13)}\]
The displacement of the particle from \(t_1\) to \(t_2\), shown by PQ, i.e. in time \(t = t_2 - t_1\), is given by
\[\vec{\Delta r} = \vec{r}_2 - \vec{r}_1 = \left(x_2 - x_1\right)\hat{i} + \left(y_2 - y_1\right)\hat{j} \qquad \text{--- (3.14)}\]
Notice how the subtraction happens separately along each axis. The \(x\)-components subtract to give \(\Delta x\), and the \(y\)-components subtract to give \(\Delta y\). Vectors never mix their components — this is the first hint of an idea that will run through the whole chapter.
Average Velocity in a Plane
We can write the average velocity of the object as
\[\vec{v}_{av} = \frac{\vec{\Delta r}}{\Delta t} = \left(\frac{x_2 - x_1}{t_2 - t_1}\right)\hat{i} + \left(\frac{y_2 - y_1}{t_2 - t_1}\right)\hat{j}\]
\[\vec{v}_{av} = \left(v_{av}\right)_x \hat{i} + \left(v_{av}\right)_y \hat{j} \qquad \text{--- (3.15)}\]
where
\[\left(v_{av}\right)_x = (x_2 - x_1)/(t_2 - t_1) \quad \text{and} \quad \left(v_{av}\right)_y = (y_2 - y_1)/(t_2 - t_1) \qquad \text{--- (3.16)}\]
Both components have the SI unit m/s. \((v_{av})_x\) tells you how fast the object drifted sideways, and \((v_{av})_y\) how fast it drifted up or down — each computed as if the other direction did not exist.
Average velocity is a vector whose direction is along \(\vec{\Delta r}\) (see Eq. (3.2)), i.e. along the direction of displacement. Dividing a vector by the scalar \(\Delta t\) (which is always positive) shrinks or stretches it but never turns it, so \(\vec{v}_{av}\) must point exactly where the displacement points.
In terms of its components, the magnitude (\(v\)) and direction (the angle \(\theta\) that the velocity vector makes with the \(x\)-axis) can be written as (see Chapter 2)
\[v_{av} = \sqrt{\left(v_{av}\right)_x^{2} + \left(v_{av}\right)_y^{2}} \quad \text{and} \quad \tan\theta = \left(v_{av}\right)_y / \left(v_{av}\right)_x \qquad \text{--- (3.17)}\]
These two relations are nothing but Pythagoras’ theorem and basic trigonometry applied to the right-angled triangle whose two perpendicular sides are the components of the vector.
Instantaneous Velocity in a Plane
Figure 3.8(b) shows the trajectory of an object moving in two dimensions. The instantaneous velocity of the object at point P along the trajectory is along the tangent to the curve at P. This is shown by the vector PQ. Its \(x\) and \(y\) components \(v_x\) and \(v_y\) are also shown in the figure.
Diagram
Fig. 3.8(b) – Instantaneous velocity: a curved trajectory with the velocity vector V drawn along the tangent at point P, resolved into its components Vₓ (horizontal) and V_y (vertical).
Why must the instantaneous velocity be along the tangent? Because as the time interval shrinks, the chord PQ becomes shorter and shorter, and a shrinking chord of a curve settles down onto the tangent at that point. This is the same limiting argument you met with the \(x\)-\(t\) graph, now applied to the path itself.
The instantaneous velocity of the object can be written in terms of derivative as (see Eq. (3.3))
\[\vec{v} = \lim_{\Delta t \to 0}\left(\frac{\vec{\Delta r}}{\Delta t}\right) = \frac{d\vec{r}}{dt} = \left(\frac{dx}{dt}\right)\hat{i} + \left(\frac{dy}{dt}\right)\hat{j} \qquad \text{--- (3.18)}\]
So the \(x\)-component of the instantaneous velocity is \(v_x = dx/dt\) and its \(y\)-component is \(v_y = dy/dt\), both in m/s. To differentiate a vector, you simply differentiate each of its components — the unit vectors \(\hat{i}\) and \(\hat{j}\) are fixed in direction, so they pass through the derivative untouched.
The magnitude and direction of the instantaneous velocity are given by
\[v = \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}} \qquad \text{--- (3.19)}\]
\[\tan\theta = \left(dy/dt\right)/\left(dx/dt\right) = dy/dx \qquad \text{--- (3.20)}\]
which is the slope of the tangent to the curve at the point at which we are calculating the instantaneous velocity.
Equation (3.20) deserves a second look. The two \(dt\) terms cancel, leaving \(dy/dx\) — a quantity that involves only the shape of the path, not the timing. That is why the direction of motion at any point depends only on where the curve is heading there, while the speed depends on how quickly the object is being carried along it.
3.3.2 Average and Instantaneous Acceleration
Again, the definitions are the same as those for rectilinear motion. Acceleration is still the rate of change of velocity — the only new feature is that the change in velocity, being a vector change, has to be tracked separately along \(x\) and along \(y\).
Average Acceleration
Thus, the average acceleration (\(\vec{a}_{av}\)) of a particle between times \(t_1\) and \(t_2\) can be written as
\[\vec{a}_{av} = \frac{\vec{v}_2 - \vec{v}_1}{t_2 - t_1} = \left(\frac{v_{2x} - v_{1x}}{t_2 - t_1}\right)\hat{i} + \left(\frac{v_{2y} - v_{1y}}{t_2 - t_1}\right)\hat{j} \qquad \text{--- (3.21)}\]
where \(\vec{v}_2\) and \(\vec{v}_1\) are the velocities of the particle at times \(t_2\) and \(t_1\) respectively. Here \(v_{1x}, v_{2x}\) are the \(x\)-components and \(v_{1y}, v_{2y}\) the \(y\)-components of these velocities, all in m/s, while the acceleration components are in m/s².
\[\vec{a}_{av} = \left(a_{av}\right)_x \hat{i} + \left(a_{av}\right)_y \hat{j} \qquad \text{--- (3.22)}\]
\(\left(a_{av}\right)_x\) and \(\left(a_{av}\right)_y\) being the \(x\) and \(y\) components of the average acceleration.
An important point hides in that definition. The average acceleration points along the change in velocity, not along the velocity itself. A car turning a corner at constant speed has a velocity pointing forward but an acceleration pointing sideways, because sideways is the direction in which its velocity is being altered.
The magnitude and direction of the acceleration are given by
\[a_{av} = \sqrt{\left(a_{av}\right)_x^{2} + \left(a_{av}\right)_y^{2}} \qquad \text{--- (3.23)}\]
and
\[\tan\theta = \left(a_{av}\right)_y / \left(a_{av}\right)_x \qquad \text{--- (3.24)}\]
which is again just Pythagoras’ theorem and trigonometry applied to the two perpendicular components.
Instantaneous Acceleration
The instantaneous acceleration is given by (see Eq. (3.5))
\[\vec{a} = \lim_{\Delta t \to 0}\left(\frac{\Delta\vec{v}}{\Delta t}\right) = \frac{d\vec{v}}{dt} = \left(\frac{dv_x}{dt}\right)\hat{i} + \left(\frac{dv_y}{dt}\right)\hat{j} \qquad \text{--- (3.25)}\]
Since \(v_x = dx/dt\) and \(v_y = dy/dt\), we can differentiate once more and write the acceleration directly in terms of the coordinates:
\[= \frac{d}{dt}\left(\frac{dx}{dt}\right)\hat{i} + \frac{d}{dt}\left(\frac{dy}{dt}\right)\hat{j} = \left(\frac{d^{2}x}{dt^{2}}\right)\hat{i} + \left(\frac{d^{2}y}{dt^{2}}\right)\hat{j} \qquad \text{--- (3.26)}\]
Thus, the \(x\) and \(y\) components of the instantaneous acceleration are respectively given by
\[a_x = d^{2}x/dt^{2} \quad \text{and} \quad a_y = d^{2}y/dt^{2} \qquad \text{--- (3.27)}\]
Both are measured in m/s². A “second derivative” simply means differentiating twice: once to get velocity from position, once more to get acceleration from velocity.
The magnitude and direction of the instantaneous acceleration are given by
\[a = \sqrt{\left(\frac{d^{2}x}{dt^{2}}\right)^{2} + \left(\frac{d^{2}y}{dt^{2}}\right)^{2}} \qquad \text{--- (3.28)}\]
and
\[\tan\theta = \left(dv_y/dt\right)/\left(dv_x/dt\right) = dv_y/dv_x \qquad \text{--- (3.29)}\]
which is the slope of the tangent to the curve in the velocity graph, i.e. a plot of \(v_y\) versus \(v_x\).
Equation (3.29) is the exact counterpart of Eq. (3.20). There, the slope of the path (\(y\) against \(x\)) gave the direction of the velocity. Here, the slope of the velocity curve (\(v_y\) against \(v_x\)) gives the direction of the acceleration.
3.3.3 Equations of Motion for an Object Travelling in a Plane with Uniform Acceleration
We have derived equations of motion for an object in rectilinear motion in Section 3.2. We will now derive similar equations for a particle moving with uniform acceleration in two dimensions.
Let the initial velocity of the object be \(\vec{u}\) at \(t = 0\) and its velocity at time \(t\) be \(\vec{v}\). As the acceleration is constant, the average acceleration and the instantaneous acceleration will be equal. (If a quantity never changes, its average over any interval is simply its constant value — so there is no need to distinguish between the two here.)
By using the definition of acceleration (Eq. (3.21)), we get
\[\vec{a} = \left(\vec{v} - \vec{u}\right)/\left(t - 0\right)\]
\[\text{or} \quad \vec{v} = \vec{u} + \vec{a}\,t \qquad \text{--- (3.30)}\]
which is the same as Eq. (3.7) but is in vector form. Here \(\vec{u}\) and \(\vec{v}\) are in m/s, \(\vec{a}\) is in m/s² and \(t\) is in seconds.
Let the displacement from time \(t = 0\) to \(t\) be \(\vec{s}\). This can be calculated from the average velocity of the object during this time. For constant acceleration, the velocity changes at a steady rate, so the average velocity is simply the mean of the initial and final velocities:
\[\vec{v}_{av} = \frac{\vec{u} + \vec{v}}{2}\]
\[\therefore \vec{s} = \left(\vec{v}_{av}\right)t = \left(\frac{\vec{u} + \vec{v}}{2}\right)t = \left(\frac{\vec{u} + \vec{u} + \vec{a}t}{2}\right)t\]
\[\therefore \vec{s} = \vec{u}\,t + \frac{1}{2}\vec{a}\,t^2 \qquad \text{--- (3.31)}\]
which is the vector form of Eq. (3.8). In the middle step we simply replaced \(\vec{v}\) by \(\vec{u} + \vec{a}t\) using Eq. (3.30).
Resolving into Components
Eq. (3.30) and (3.31) can be resolved into their \(x\) and \(y\) components so as to get corresponding scalar equations as follows.
\[v_x = u_x + a_x t \qquad \text{--- (3.32)}\]
and
\[v_y = u_y + a_y t \qquad \text{--- (3.33)}\]
\[s_x = u_x t + \frac{1}{2}a_x t^{2} \qquad \text{--- (3.34)}\]
and
\[s_y = u_y t + \frac{1}{2}a_y t^{2} \qquad \text{--- (3.35)}\]
A vector equation is true only if it is true separately for each component, because \(\hat{i}\) and \(\hat{j}\) point in genuinely different directions and can never substitute for one another. That is why one vector equation splits cleanly into two scalar equations.
Now look carefully at what these four equations contain. We can see that Eqs. (3.32) and (3.34) involve only the \(x\) components of displacement, velocity and acceleration, while Eqs. (3.33) and (3.35) involve only the \(y\) components of these quantities.
Thus the two sets of equations are independent of each other and can be solved independently. Nothing in the \(x\)-equations refers to \(y\), and nothing in the \(y\)-equations refers to \(x\).
We can thus see that the motion along the \(x\) direction of an object is completely controlled by the \(x\) components of velocity and acceleration, while that along the \(y\) direction is completely controlled by the \(y\) components of these quantities. This makes it easy to study the motion in two dimensions, which gets converted to two independent rectilinear motions along two perpendicular directions.
3.3.4 Relative Velocity
Relative velocity between two objects moving in a plane can be defined in a way similar to that for objects moving along a straight line. The idea is unchanged — what you observe from a moving platform is the difference of the two velocities — but the subtraction must now be carried out as a vector subtraction, component by component.
The relative velocity of object A having velocity \(\vec{v}_A\), with respect to the object B having velocity \(\vec{v}_B\), is given by
\[\vec{v}_{AB} = \vec{v}_A - \vec{v}_B \qquad \text{--- (3.36)}\]
Similarly, the relative velocity of object B with respect to object A is given by
\[\vec{v}_{BA} = \vec{v}_B - \vec{v}_A \qquad \text{--- (3.37)}\]
We can see that the magnitudes of the two relative velocities (\(v_{AB}\) and \(v_{BA}\)) are equal and their directions are opposite.
Diagram
Fig. 3.9 – Two velocity vectors v_A and v_B drawn from a common point, with the vector v_A – v_B constructed by reversing v_B and adding it to v_A, giving the relative velocity v_AB.
Chaining Relative Velocities
Consider a number of objects A, B, C, D —- Y, Z, moving with respect to the other. Using the symbol \(\vec{v}_{AB}\) for representing the velocity of A relative to B, etc., the velocity of A relative to Z can be written as
\[\vec{v}_{AZ} = \vec{v}_{AB} + \vec{v}_{BC} + \vec{v}_{CD} + \ldots + \vec{v}_{XY} + \vec{v}_{YZ}\]
Note the order of subscripts (A→B→C→D— →Z).
The rule is easy to remember if you read the subscripts as a chain: the second letter of each term must match the first letter of the next term, and the letters in the middle simply cancel out, leaving only the first and the last. A useful consequence, used constantly in problems, is \(\vec{v}_{AB} = \vec{v}_{AC} + \vec{v}_{CB}\) — you may pass through any convenient intermediate object.
3.3.5 Projectile Motion
Any object in flight after being thrown with some velocity is called a projectile, and its motion is called projectile motion.
We often see projectile motion in our day-to-day life. Children throw stones towards trees for getting tamarind pods or mangoes. A bowler bowls a ball towards a batsman in cricket; a basketball player throws a ball towards the basket. All these are illustrations of projectile motion.
In this motion, we have objects (projectiles) with a given initial velocity, moving under the influence of the Earth’s gravitational field. Note the crucial point: once the object leaves the hand (or the bat, or the barrel), nothing is pushing it forward any more. Whatever happens afterwards is decided entirely by the velocity it was given and by gravity.
The Two Components of the Motion
The projectile has two components of velocity, one in the horizontal, i.e. along the \(x\)-direction, and the other in the vertical, i.e. along the \(y\) direction. The acceleration due to gravity acts only along the vertically downward direction.
The horizontal component of velocity, therefore, remains unchanged, as no force is acting in the horizontal direction, while the vertical component changes in accordance with the laws of motion, with \(\vec{a}_x\) being 0 and \(\vec{a}_y\) (\(= -\vec{g}\)) being the downward acceleration due to gravity (upward is positive).
Unless stated otherwise, retarding forces like air resistance, etc., are neglected for the projectile motion. This is the assumption that makes the whole treatment simple, and you should keep it consciously in mind.
Setting Up the Equations
Let us assume that the initial velocity of the projectile is \(\vec{u}\) and its direction makes an angle \(\theta\) with the horizontal, as shown in Fig. 3.10. The projectile is thrown from the ground. We take the \(x\)-axis along the ground and the \(y\)-axis in the vertical direction.
Diagram
Fig. 3.10 – Trajectory of a projectile launched from origin O at angle θ, showing the parabolic path through points A, P (highest point) and B, ending at Q on the ground, with velocity components uₓ and u_y marked at each point and the range R marked along the ground.
The horizontal and vertical components of initial velocity are \(u\cos\theta\) and \(u\sin\theta\) respectively. The horizontal component remains unchanged in the absence of any force acting in that direction, while the vertical component changes according to Eq. (3.33), with \(a_y = -g\) and \(u_y = u\sin\theta\).
Thus, the components of velocity at time \(t\) are given by
\[v_x = u_x = u\cos\theta \qquad \text{--- (3.38)}\]
\[v_y = u_y - gt = u\sin\theta - gt \qquad \text{--- (3.39)}\]
Here \(u\) is the initial speed in m/s, \(\theta\) the angle of projection, \(g\) the acceleration due to gravity in m/s², and \(t\) the time in seconds. As \(0 < \theta < 90°\), the vertical component initially is in the upward direction.
Similarly, the displacements of the projectile in the horizontal and vertical directions at time \(t\), according to Eqs. (3.34) and (3.35), are given by
\[s_x = u\cos\theta \cdot t \qquad \text{--- (3.40)}\]
\[s_y = u\sin\theta \cdot t - \frac{1}{2}gt^2 \qquad \text{--- (3.41)}\]
Notice that Eq. (3.40) has no \(t^2\) term at all. That is the mathematical signature of zero horizontal acceleration: the horizontal displacement simply grows in proportion to time, exactly as it would for an object drifting freely with nothing acting on it.
The direction of motion of the projectile at any time \(t\) makes an angle \(\alpha\) with the horizontal, which is given by
\[\tan\alpha = v_y(t)/v_x(t) \qquad \text{--- (3.42)}\]
Since \(v_x\) stays fixed while \(v_y\) keeps falling, this angle keeps changing throughout the flight — the projectile is continuously turning, which is another way of saying its path is curved.
Following the Projectile Through Its Flight
The vertical velocity keeps on decreasing as the projectile goes up and becomes zero at a certain time. At that time the height of the projectile is maximum. The velocity then starts increasing in the downward direction, as the particle is now falling under the Earth’s gravitational field with a constant horizontal component of velocity. After a while the projectile reaches the ground.
The trajectory of the object is shown in Fig. 3.10. The projectile is assumed to start from the origin of the coordinate system, O. The point of maximum height is indicated by P, and the point where it falls down to the ground is indicated by Q. The horizontal and vertical components of velocity are shown at these points, as well as at two intermediate points A and B on the trajectory of the projectile.
Note that the horizontal component of velocity remains the same, i.e. \(u_x\), while the vertical component decreases and becomes zero at P. After that it changes its direction, its magnitude increases, and becomes equal to \(u_y\) again at Q. The horizontal distance covered by the projectile before it falls to the ground is OQ.
We can derive the equation of the trajectory of the projectile as follows.
Time of Flight
Let the time taken by the projectile to reach the maximum height be \(t_0\). The trajectory of the object being symmetrical, it can be shown by using equations of motion that the object will take the same time in going up in the air and coming down to the ground.
At the highest point P, \(t = t_0\) and \(v_y = 0\). Using Eq. (3.39), we get,
\[0 = u\sin\theta - gt_0\]
\[t_0 = (u\sin\theta)/g \qquad \text{--- (3.43)}\]
\(\therefore\) Total time in air \(= T = 2t_0\) is the time of flight.
Look at what \(t_0\) depends on: only \(u\sin\theta\), the vertical part of the initial velocity, and \(g\). The horizontal speed plays no role in deciding how long the projectile stays up. This is the independence of the two motions (Section 3.3.3) showing itself in a very practical way.
Horizontal Range
The total horizontal distance travelled by the particle in this time \(T\) can be obtained by using Eq. (3.40) as
\[R = u_x \cdot T = u\cos\theta \cdot 2t_0 = u\cos\theta \cdot (2u\sin\theta)/g\]
\[= 2\,u_x u_y/g = u^2(2\sin\theta\cos\theta)/g\]
\[= u^2 \sin 2\theta / g \qquad \text{--- (3.44)}\]
In the last step we used the trigonometric identity \(2\sin\theta\cos\theta = \sin 2\theta\).
This maximum horizontal distance travelled by the projectile is called the horizontal range \(R\) of the projectile, and it depends on the magnitude and direction of the initial velocity of the projectile, as well as on the value of acceleration due to gravity at that place.
For maximum horizontal range,
\[\sin 2\theta = 1 \quad \therefore \ 2\theta = 90° \ or \ \theta = 45°\]
Hence,
\[R = R_{max} = \frac{u^2}{g} \quad \text{for} \ \theta = 45°\]
The reasoning is simple: in \(R = u^2\sin 2\theta/g\), the only quantity that the thrower can vary (for a fixed launch speed) is \(\sin 2\theta\), and a sine can never exceed 1. It reaches 1 when \(2\theta = 90°\), i.e. \(\theta = 45°\).
Maximum Height
The maximum height \(H\) reached by the projectile, having a certain value of \(\theta\), is the distance travelled along the vertical (\(y\)) direction in time \(t_0\). This can be calculated by using Eq. (3.41) as
\[H = u\sin\theta \cdot t_0 - \frac{1}{2}g t_0^2\]
Substituting \(t_0 = u\sin\theta/g\) from Eq. (3.43),
\[= u\sin\theta\left(\frac{u\sin\theta}{g}\right) - \frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2\]
\[= \frac{u^2\sin^2\theta}{2g} = \frac{u_y^2}{2g} \qquad \text{--- (3.45)}\]
The final form \(H = u_y^2/(2g)\) is worth memorising, because it is nothing but the third equation of motion applied to the vertical direction alone: a body launched upward with speed \(u_y\) rises until its vertical speed is used up.
Equation of Motion for a Projectile
We can derive the equation of motion of the projectile, which is the relation between the displacements of the projectile along the vertical and horizontal directions. This can be obtained by eliminating \(t\) between the equations giving these displacements, i.e. Eqs. (3.40) and (3.41).
Why eliminate \(t\)? Because Eqs. (3.40) and (3.41) tell us where the projectile is at each moment, but an equation linking \(y\) directly to \(x\) tells us the shape of the path itself, with time removed from the picture.
As the projectile starts from \(\vec{x} = 0\), we can write \(s_x = x\) and \(s_y = y\).
\[\therefore s_x = \left(u\cos\theta\right)t \quad \therefore t = \frac{s_x}{u\cos\theta} = \frac{x}{u\cos\theta}\]
\[\therefore y = \left(u\sin\theta\right)t - \frac{1}{2}gt^2\]
\[= \left(u\sin\theta\right)\left(\frac{x}{u\cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2\]
\[\therefore y = \left(\tan\theta\right)x - \frac{1}{2}\left(\frac{g}{u^2\cos^2\theta}\right)x^2 \qquad \text{--- (3.46)}\]
This is the equation of the trajectory of the projectile. Here, \(u\) and \(\theta\) are constants for the given projectile motion.
The above equation is of the form
\[y = Ax + Bx^2 \qquad \text{--- (3.47)}\]
with \(A = \tan\theta\) and \(B = -g/(2u^2\cos^2\theta)\), both fixed numbers once \(u\) and \(\theta\) are chosen. This is the equation of a parabola. Thus, the path, i.e. the trajectory, of a projectile is a parabola.
Diagram
Fig. 3.11 – A cone cut by a plane parallel to its slant side, producing a parabola; alongside it, a parabola with its focus, directrix and latus rectum labelled.
3.4 Uniform Circular Motion
An object moving with constant speed along a circular path is said to be in uniform circular motion (UCM). Such a motion is only possible if its velocity is always tangential to its circular path, without change in its magnitude.
Read the definition carefully. The speed is constant, but the velocity is not — because the direction of the tangent keeps swinging around as the object goes round the circle. In UCM, the magnitude of the velocity never changes, while its direction changes continuously.
Diagram
Fig. 3.12 – A particle moving anticlockwise along a circle, with velocity vectors drawn tangentially at four different positions, showing equal lengths but different directions.
Why an Acceleration Is Needed
To change the direction of velocity, acceleration is a must. Acceleration, after all, is nothing but the rate of change of velocity, and any change — in magnitude or in direction — requires it.
However, if the acceleration or its component is in line with the velocity (along or opposite to the velocity), it will always change the speed (magnitude of velocity), in which case the object will not continue its uniform circular motion.
This gives us a strict condition. We need an acceleration that turns the velocity without ever speeding it up or slowing it down. The only way to achieve that is to allow the acceleration no component along the velocity at all — that is, to keep it exactly perpendicular to the velocity.
In order to achieve both these requirements, the acceleration must be (i) perpendicular to the tangential velocity, (ii) of constant magnitude, and (iii) always directed towards the centre of the circular trajectory.
Such an acceleration is called centripetal (centre-seeking) acceleration, and the force causing this acceleration is centripetal force.
The Two Requirements for Circular Motion
Thus, in order to realise a circular motion, there are two requirements: (i) tangential velocity, and (ii) centripetal force of suitable constant magnitude.
Both requirements are essential, and removing either one destroys the motion. With no tangential velocity, the object would simply be pulled straight in towards the centre. With no centripetal force, it would fly off along the tangent in a straight line, exactly as Newton’s first law demands.
An example is the motion of the moon going around the Earth in a nearly circular orbit, as a result of the constant gravitational attraction of fixed magnitude felt by it towards the Earth. The moon has a tangential velocity, and the Earth’s gravity supplies the inward pull — the two requirements together keep it in orbit instead of letting it either fall in or escape.
3.4.1 Period, Radius Vector and Angular Speed
Consider an object of mass \(m\), moving with a uniform speed \(v\), along a circle of radius \(r\). Let \(T\) be the time period of revolution of the object, i.e. the time taken by the object to complete one revolution, or to travel a distance of \(2\pi r\).
The distance \(2\pi r\) is simply the circumference of the circle — the full path length covered in one round.
Since the speed is uniform, the time taken is just the distance divided by the speed. Thus,
\[T = 2\pi r / v\]
Rearranging the same relation gives the speed:
\[\therefore \text{Speed } v = \frac{\text{Distance}}{\text{Time}} = \frac{2\pi r}{T} \qquad \text{--- (3.48)}\]
where \(r\) is the radius in metres, \(T\) is the period in seconds, and \(v\) is the speed in m/s. This speed \(v\), being directed along the tangent, is often called the linear speed (or tangential speed) of the particle.
The Radius Vector
During circular motion of a point object, the position vector of the object from the centre of the circle is the radius vector \(\vec{r}\). Its magnitude is the radius \(r\), and it is directed away from the centre to the particle, i.e. away from the centre of the circle.
Diagram
Fig. 3.13 – Uniform circular motion: a particle P moving anticlockwise along a circle of radius r centred at the origin O, starting from P₀ on the +x axis, with the radius vector r making an angle θ = ωt with the x-axis.
The magnitude of \(\vec{r}\) never changes during the motion — a circle is, by definition, the set of points at a fixed distance from the centre. What does change, continuously, is the direction of \(\vec{r}\). As the particle performs UCM, this radius vector describes equal angles in equal intervals of time.
Angular Speed
At this stage we can define a new quantity called angular speed \(\omega\), which gives the angle described by the radius vector, per unit time. It is analogous to speed, which is distance travelled per unit time.
During one complete revolution, the angle described is \(2\pi\) and the time taken is the period \(T\). Hence, the angular speed
\[\omega = \frac{\text{Angle}}{\text{time}} = \frac{2\pi}{T} = \frac{\left(2\pi\right)}{\left(\dfrac{2\pi r}{\text{v}}\right)} = \frac{\text{v}}{r} \qquad \text{--- (3.49)}\]
The unit of \(\omega\) is radian/sec.
In the third step we simply substituted \(T = 2\pi r/v\) from Eq. (3.48); the \(2\pi\) cancels, and we are left with the compact and very useful result \(v = \omega r\).
3.4.2 Expression for Centripetal Acceleration
Figure 3.13 shows a particle P performing a UCM in an anticlockwise sense along a circle of radius \(r\) with angular speed \(\omega\) and period \(T\). Let us choose the coordinates such that this motion is in the \(xy\)-plane, having centre at the origin O.
Initially (for simplicity), let the particle be at \(P_0\) on the positive \(x\)-axis. At a given instant \(t\), the radius vector of P makes an angle \(\theta\) with the \(x\)-axis.
Since the radius vector sweeps out equal angles in equal times, the angle grows in direct proportion to time:
\[\therefore \theta = \omega t \quad \text{and so} \quad \frac{d\theta}{dt} = \omega\]
\(x\) and \(y\) components of the radius vector \(\vec{r}\) will then be \(r\cos\theta\) and \(r\sin\theta\) respectively — the standard resolution of a vector of magnitude \(r\) making an angle \(\theta\) with the \(x\)-axis.
\[\therefore \vec{r} = \left(r\cos\theta\right)\hat{i} + \left(r\sin\theta\right)\hat{j}\]
\[= \left(r\cos\left[\omega t\right]\right)\hat{i} + \left(r\sin\left[\omega t\right]\right)\hat{j} \qquad \text{--- (3.50)}\]
This single equation contains the whole of uniform circular motion. Everything else now follows by differentiation.
Differentiating to Get Velocity and Acceleration
Time derivative of position vector \(\vec{r}\) gives instantaneous velocity \(\vec{v}\), and time derivative of velocity \(\vec{v}\) gives instantaneous acceleration \(\vec{a}\). Magnitudes of \(r\) and \(\omega\) are constants, so they come out of the derivative; only the sine and cosine terms are differentiated.
\[\therefore \vec{v} = \frac{d\vec{r}}{dt} = r\left(-\omega\sin\left[\omega t\right]\hat{i} + \omega\cos\left[\omega t\right]\hat{j}\right)\]
\[= r\omega\left(-\sin\left[\omega t\right]\hat{i} + \cos\left[\omega t\right]\hat{j}\right) \qquad \text{--- (3.51)}\]
Differentiating once more,
\[\therefore \vec{a} = \frac{d\vec{v}}{dt} = r\omega\left(-\omega\cos\left[\omega t\right]\hat{i} - \omega\sin\left[\omega t\right]\hat{j}\right)\]
\[= -\omega^2\left(r\cos\left[\omega t\right]\hat{i} + r\sin\left[\omega t\right]\hat{j}\right) = -\omega^2 \vec{r} \qquad \text{--- (3.52)}\]
In the last step we recognised the bracket as exactly the position vector \(\vec{r}\) of Eq. (3.50).
Here the minus sign shows that the acceleration is opposite to that of \(\vec{r}\), i.e. towards the centre. This is the centripetal acceleration.
This is a remarkable result: the mathematics itself, with no extra assumption, has produced an acceleration pointing straight at the centre — precisely the inward acceleration that Section 3.4 argued was necessary.
The magnitude of acceleration,
\[a = \omega^2 r = \frac{v^2}{r} = \omega v \qquad \text{--- (3.53)}\]
Here \(\omega\) is the angular speed in rad/s, \(r\) the radius in m, \(v\) the linear speed in m/s, and \(a\) the centripetal acceleration in m/s². The three forms are equivalent, obtained from one another using \(v = \omega r\); choose whichever matches the data you are given.
The force providing this acceleration should also be along the same direction, hence centripetal.
\[\therefore \vec{F} = m\vec{a} = -m\omega^2 \vec{r} \qquad \text{--- (3.54)}\]
Magnitude of
\[F = m\omega^2 r = \frac{m v^2}{r} = m\omega v \qquad \text{--- (3.55)}\]
with \(m\) in kg and \(F\) in newtons (N).
Conical Pendulum
In a simple pendulum, a mass \(m\) is suspended by a string of length \(l\) and moves along an arc of a vertical circle. If the mass instead revolves in a horizontal circle, and the string, which makes a constant angle with the vertical, describes a cone whose vertex is the fixed point O, then the mass-string system is called a conical pendulum, as shown in Fig. 3.14. In the absence of friction, the system will continue indefinitely once started.
Diagram
Fig. 3.14 – Conical pendulum: a bob of mass m hanging by a string of length l from a fixed point O, revolving in a horizontal circle of radius r = l sinθ at a depth h = l cosθ below O, with the tension T resolved into T cosθ (vertical) and T sinθ (horizontal) and the weight mg shown downward.
As shown in the figure, the forces acting on the bob of mass \(m\) of the conical pendulum are: (i) Gravitational force, \(mg\), acting vertically downwards, and (ii) Force due to tension \(\vec{T}\), acting along the string, directed towards the support. These are the only two forces acting on the bob.
For the bob to undergo horizontal circular motion (on a circle of radius \(r\)), the resultant force must be centripetal (directed towards the centre of the circle). In other words, the vertical gravitational force must be balanced.
Thus, we resolve tension \(\vec{T}\) into two mutually perpendicular components. Let \(\theta\) be the angle made by the string with the vertical at any position. The component \(T\cos\theta\) is acting vertically upwards. The inclination should be such that
\[T\cos\theta = mg,\]
so that there is no net vertical force. (This must be so: the bob stays at the same height throughout, so it has no vertical acceleration, and hence no net vertical force.)
The resultant force on the bob is then \(T\sin\theta\), which is radial, or centripetal, or directed towards the centre O′.
\[\therefore \ T\sin\theta = \frac{mv^2}{r}\]
Dividing the horizontal equation by the vertical one, the tension \(T\) cancels out — a standard and very useful trick, because \(T\) is usually the quantity we do not know:
\[\tan\theta = \frac{T\sin\theta}{T\cos\theta} = \frac{\left(mv^2/r\right)}{mg} = \frac{v^2}{rg}\]
The mass \(m\) also cancels here. Keep that in mind; it will matter in a moment.
Since we know \(v = \dfrac{2\pi r}{T}\) (from Eq. (3.48), where here \(T\) denotes the time period of one revolution),
\[\therefore \tan\theta = \frac{4\pi^2 r^2}{T^2 r g}\]
\[T = 2\pi \sqrt{\frac{r}{g\tan\theta}}\]
\[T = 2\pi \sqrt{\frac{l\sin\theta}{g\tan\theta}} \qquad \left(\because r = l\sin\theta\right)\]
Using \(\tan\theta = \sin\theta/\cos\theta\), the \(\sin\theta\) cancels and leaves a \(\cos\theta\):
\[T = 2\pi \sqrt{\frac{l\cos\theta}{g}}\]
\[T = 2\pi \sqrt{\frac{h}{g}} \qquad \left(\because h = l\cos\theta\right) \qquad \text{--- (3.56)}\]
where \(l\) is the length of the pendulum and \(h\) is the vertical distance of the horizontal circle from the fixed point O. Here \(T\) is the time period in seconds, \(l\) and \(h\) are in metres and \(g\) is in m/s².
3.5 Summary
Kinds of Motion
- Motion is a change in the position of an object with time.
- Motion is of three types: rectilinear motion (along a straight line), motion in a plane (two dimensions), and motion in space (three dimensions).
- In rectilinear motion, force and velocity lie along the same line, so direction can be shown simply by a \(+ve\) or \(-ve\) sign. In a plane, they generally point in different directions, so vectors must be used.
Basic Quantities of Motion
- Displacement: \(\vec{s} = \Delta\vec{x} = \vec{x}_2 - \vec{x}_1\) — (3.1). A vector; SI unit m; dimensions \([\mathrm{L^1M^0T^0}]\).
- Path length: the actual distance covered along the path. A scalar; SI unit m. Its magnitude is never less than the magnitude of displacement.
- Average velocity: \(\vec{v}_{av} = \dfrac{\vec{x}_2 - \vec{x}_1}{(t_2 - t_1)}\) — (3.2). A vector along the displacement; SI unit m/s; dimensions \([\mathrm{L^1M^0T^{-1}}]\).
- Average speed \(= \dfrac{\text{path length}}{\text{time interval}}\). A scalar with the same dimensions as velocity.
- Instantaneous velocity: \(\vec{v} = \lim\limits_{\Delta t \to 0}\left(\dfrac{\Delta\vec{x}}{\Delta t}\right) = \dfrac{d\vec{x}}{dt}\) — (3.3).
- Instantaneous speed is always equal to the magnitude of the instantaneous velocity.
- Acceleration: \(\vec{a} = \dfrac{\left(\vec{v}_2 - \vec{v}_1\right)}{\left(t_2 - t_1\right)}\) — (3.4), and \(\vec{a} = \lim\limits_{\Delta t \to 0}\left(\dfrac{\Delta\vec{v}}{\Delta t}\right) = \dfrac{d\vec{v}}{dt}\) — (3.5). A vector; SI unit m/s²; dimensions \([\mathrm{L^1M^0T^{-2}}]\).
Graphs of Motion
- The slope of the position-time (\(x\)-\(t\)) graph gives the velocity; the slope of a chord gives average velocity, and the slope of the tangent gives instantaneous velocity.
- The slope of the velocity-time (\(v\)-\(t\)) graph gives the acceleration.
- The area under the \(v\)-\(t\) graph gives the displacement: \[\text{Area} = \int_{t_1}^{t_2} v\,dt = \int_{t_1}^{t_2}\frac{dx}{dt}\,dt = x(t_2) - x(t_1) \qquad \text{--- (3.6)}\]
- While using the area-under-the-curve method, the origin of the velocity axis must be zero.
Equations of Motion (Uniform Acceleration, Rectilinear)
- \(v = u + at\) — (3.7)
- \(s = ut + \dfrac{1}{2}at^2\) — (3.8)
- \(v^2 - u^2 = 2a\cdot s\) — (3.9)
These hold only when the acceleration is uniform.
- Free fall: a body starting from rest at some height and falling only under the Earth’s gravity. Its acceleration is \(g\), along the vertical, and may be taken as constant over distances small compared with the Earth’s radius.
- For a body starting from rest with uniform acceleration, the distances covered in successive equal time intervals are in the ratio 1 : 3 : 5 : 7 …
Relative Velocity
- \(v_{AB} = v_A - v_B\) — (3.10) and \(v_{BA} = v_B - v_A\) — (3.11).
- In a plane: \(\vec{v}_{AB} = \vec{v}_A - \vec{v}_B\) — (3.36) and \(\vec{v}_{BA} = \vec{v}_B - \vec{v}_A\) — (3.37). The two have equal magnitudes and opposite directions.
- Chain rule: \(\vec{v}_{AZ} = \vec{v}_{AB} + \vec{v}_{BC} + \vec{v}_{CD} + \ldots + \vec{v}_{XY} + \vec{v}_{YZ}\), keeping the order of subscripts A→B→C→ … →Z.
Motion in a Plane
- Position vector: \(\vec{r}_1 = x_1\hat{i} + y_1\hat{j}\) — (3.12); displacement \(\vec{\Delta r} = \left(x_2 - x_1\right)\hat{i} + \left(y_2 - y_1\right)\hat{j}\) — (3.14).
- Average velocity: \(\vec{v}_{av} = \left(v_{av}\right)_x\hat{i} + \left(v_{av}\right)_y\hat{j}\) — (3.15), with magnitude \(v_{av} = \sqrt{\left(v_{av}\right)_x^2 + \left(v_{av}\right)_y^2}\) and \(\tan\theta = \left(v_{av}\right)_y/\left(v_{av}\right)_x\) — (3.17).
- Instantaneous velocity: \(\vec{v} = \dfrac{d\vec{r}}{dt} = \left(\dfrac{dx}{dt}\right)\hat{i} + \left(\dfrac{dy}{dt}\right)\hat{j}\) — (3.18); \(\tan\theta = dy/dx\) — (3.20), the slope of the tangent to the path.
- Instantaneous acceleration: \(\vec{a} = \dfrac{d\vec{v}}{dt} = \left(\dfrac{d^2x}{dt^2}\right)\hat{i} + \left(\dfrac{d^2y}{dt^2}\right)\hat{j}\) — (3.25), (3.26); so \(a_x = d^2x/dt^2\) and \(a_y = d^2y/dt^2\) — (3.27).
Equations of Motion in a Plane (Uniform Acceleration)
- \(\vec{v} = \vec{u} + \vec{a}t\) — (3.30) and \(\vec{s} = \vec{u}t + \dfrac{1}{2}\vec{a}t^2\) — (3.31).
- Resolved into components:
- \(v_x = u_x + a_x t\) — (3.32); \(\ v_y = u_y + a_y t\) — (3.33)
- \(s_x = u_x t + \dfrac{1}{2}a_x t^2\) — (3.34); \(\ s_y = u_y t + \dfrac{1}{2}a_y t^2\) — (3.35)
- The \(x\)-equations and \(y\)-equations are completely independent of each other. Motion in two dimensions can be resolved into two independent motions in mutually perpendicular directions.
Projectile Motion
For a projectile launched from the ground with speed \(u\) at an angle \(\theta\) to the horizontal, with \(a_x = 0\) and \(a_y = -g\), and air resistance neglected:
- \(v_x = u_x = u\cos\theta\) — (3.38); \(\ v_y = u\sin\theta - gt\) — (3.39)
- \(s_x = u\cos\theta \cdot t\) — (3.40); \(\ s_y = u\sin\theta \cdot t - \dfrac{1}{2}gt^2\) — (3.41)
- Direction of motion at time \(t\): \(\tan\alpha = v_y(t)/v_x(t)\) — (3.42)
- Time to reach the highest point: \(t_0 = (u\sin\theta)/g\) — (3.43); time of flight \(T = 2t_0\).
- Horizontal range: \(R = u^2\sin 2\theta/g\) — (3.44). It is maximum for \(\theta = 45°\), where \(R_{max} = \dfrac{u^2}{g}\).
- Maximum height: \(H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{u_y^2}{2g}\) — (3.45)
- Trajectory: \(y = \left(\tan\theta\right)x - \dfrac{1}{2}\left(\dfrac{g}{u^2\cos^2\theta}\right)x^2\) — (3.46), which is of the form \(y = Ax + Bx^2\) — (3.47). Hence the path of a projectile is a parabola.
- These expressions for \(T\), \(R\), \(R_{max}\) and \(H\) hold only when air resistance is absent. In reality, the time of ascent \(t_a\) is greater than the time of descent \(t_d\), the maximum range occurs at an angle greater than 45°, and the range is much less than \(u^2/g\).
Uniform Circular Motion
- In UCM, the speed is constant and the velocity is always tangential to the circle; only the direction of velocity changes.
- Two requirements: a tangential velocity, and a centripetal force of suitable constant magnitude directed towards the centre.
- Period and speed: \(T = 2\pi r/v\), i.e. \(v = \dfrac{2\pi r}{T}\) — (3.48).
- Angular speed: \(\omega = \dfrac{2\pi}{T} = \dfrac{v}{r}\) — (3.49). SI unit: radian/second.
- Position vector: \(\vec{r} = \left(r\cos\left[\omega t\right]\right)\hat{i} + \left(r\sin\left[\omega t\right]\right)\hat{j}\) — (3.50)
- Velocity: \(\vec{v} = r\omega\left(-\sin\left[\omega t\right]\hat{i} + \cos\left[\omega t\right]\hat{j}\right)\) — (3.51)
- Centripetal acceleration: \(\vec{a} = -\omega^2\vec{r}\) — (3.52), of magnitude \(a = \omega^2 r = \dfrac{v^2}{r} = \omega v\) — (3.53). The minus sign shows it is directed towards the centre.
- Centripetal force: \(\vec{F} = m\vec{a} = -m\omega^2\vec{r}\) — (3.54), of magnitude \(F = m\omega^2 r = \dfrac{mv^2}{r} = m\omega v\) — (3.55).
- Centripetal force is not an extra external force; it is the resultant of the real forces acting on the body. In planetary motion, gravitation provides it.
Conical Pendulum
- A bob revolving in a horizontal circle on a string that sweeps out a cone.
- The forces on the bob are only the weight \(mg\) and the tension \(\vec{T}\). Vertically, \(T\cos\theta = mg\); horizontally, \(T\sin\theta = \dfrac{mv^2}{r}\), so \(\tan\theta = \dfrac{v^2}{rg}\).
- Time period: \(T = 2\pi\sqrt{\dfrac{l\cos\theta}{g}} = 2\pi\sqrt{\dfrac{h}{g}}\) — (3.56), where \(l\) is the length of the string, \(\theta\) the angle with the vertical, and \(h = l\cos\theta\) the vertical depth of the circle below the point of support. The period does not depend on the mass of the bob.
3.6 Check Your Concepts
Why can the magnitude of displacement never exceed the path length, no matter how the object moves?
An athlete completes exactly one lap of a circular track and returns to the starting line. Her average speed is 5 m/s, but her average velocity is zero. Explain how both statements can be true at the same time.
Average speed and the magnitude of average velocity may differ, but instantaneous speed and the magnitude of instantaneous velocity are always equal. Give the reason.
What does the slope of a chord of a position-time graph represent, and what does the slope of the tangent at a point on the same graph represent? Explain how one becomes the other.
A student uses \(v = u + at\) to solve a problem in which the acceleration increases steadily with time. Why is this wrong, and what should be used instead?
Why is the area under a velocity-time graph equal to the displacement and not to the path length? Under what condition do the two become equal?
Two cars travel towards each other along a straight road, each at 40 km/hr with respect to the ground. Explain, using the definition of relative velocity, why the driver of one car sees the other approaching at 80 km/hr.
Why must vector equations be used for motion in a plane, whereas signs alone were sufficient for rectilinear motion?
A stone is dropped from a height and, at the same instant, another stone is thrown horizontally from the same height. Which stone reaches the ground first? Justify your answer using the equations of motion in a plane.
At the highest point of its path, is the velocity of a projectile zero? What about its acceleration? Explain both.
For a given launch speed, why does a projectile achieve its maximum horizontal range at 45°, and why is the actual optimum angle greater than 45° in real conditions where air resistance is present?
Show, from Eqs. (3.40) and (3.41), why eliminating \(t\) leads to an equation of the form \(y = Ax + Bx^2\), and state what this tells you about the shape of a projectile’s path.
An object moves along a circle with constant speed. Its velocity is changing, yet its speed is not. Explain this apparent contradiction, and state the direction of its acceleration.
Why must the acceleration in uniform circular motion be perpendicular to the velocity at every instant? What would happen if it had a component along the velocity?
If the string of a whirling stone suddenly snaps, in which direction does the stone move? Explain with reference to Newton’s first law.
“Centripetal force is not an extra external force acting on a body.” Explain this statement using the example of the conical pendulum and the example of a planet orbiting the Sun.
Two children sit on a rotating merry-go-round, one near the axis and one at the rim. Compare their angular speeds, their linear speeds and their centripetal accelerations.
The period of a conical pendulum is \(T = 2\pi\sqrt{h/g}\). Explain why a heavier bob does not take longer to complete one revolution.
Separate the following into groups of scalars and vectors: velocity, speed, displacement, work done, force, power, energy, acceleration, electric charge, angular velocity.
Define average velocity and instantaneous velocity. When are they the same?
Define free fall.
If the motion of an object is described by \(x = f(t)\), write the formulae for instantaneous velocity and acceleration.
Derive the equations of motion for a particle moving in a plane and show that the motion can be resolved into two independent motions in mutually perpendicular directions.
Derive the equations of motion graphically for a particle having uniform acceleration, moving along a straight line.
Derive the formula for the range and maximum height achieved by a projectile thrown from the origin with initial velocity \(\vec{u}\) at an angle \(\theta\) to the horizontal.
What is a conical pendulum? Show that its time period is given by \(2\pi\sqrt{\dfrac{l\cos\theta}{g}}\), where \(l\) is the length of the string, \(\theta\) is the angle that the string makes with the vertical, and \(g\) is the acceleration due to gravity.
Define angular velocity. Show that the centripetal force on a particle undergoing uniform circular motion is \(-m\omega^2\vec{r}\).
3.7 Practice with Numericals
An object moves along the \(x\)-axis from \(x = 3\) m to \(x = 11\) m in 4 s, then reverses and reaches \(x = 5\) m in the next 2 s. Calculate (i) the total path length, (ii) the displacement, (iii) the average speed, and (iv) the magnitude of the average velocity.
A motorcycle starting from rest accelerates uniformly at 2.5 m/s² for 8 s along a straight road. Find (i) its velocity at the end of 8 s, (ii) the distance covered in this time, and (iii) the distance covered in the 8th second alone.
A train moving at 20 m/s is brought to rest in 25 s by applying brakes. Assuming uniform retardation, find (i) the retardation, and (ii) the distance travelled before stopping.
A stone is dropped from rest from the top of a tower and reaches the ground in 4 s. Taking \(g = 10\) m/s², find (i) the height of the tower, (ii) the speed of the stone on striking the ground, and (iii) the ratio of the distances covered in the 1st, 2nd and 3rd seconds of its fall.
A ball is thrown vertically upward with a speed of 30 m/s. Taking \(g = 10\) m/s², find (i) the maximum height it reaches, (ii) the time taken to reach that height, and (iii) the total time it remains in the air before returning to the point of projection.
A car travels the first 20 km of a journey at 40 km/hr and the next 30 km at 60 km/hr. Find the average speed of the car for the whole journey.
Two trains move on parallel tracks in the same direction with speeds 90 km/hr and 72 km/hr with respect to the ground. Find (i) the velocity of the faster train relative to the slower one, and (ii) the time the faster train takes to completely overtake the slower one if each train is 150 m long.
The position vector of a particle moving in the \(x\)-\(y\) plane is \(\vec{r} = \left(2t^2\right)\hat{i} + \left(3t^3\right)\hat{j}\) m, with \(t\) in seconds. Determine (i) its velocity vector, (ii) its acceleration vector, and (iii) the magnitude of its acceleration at \(t = 1\) s.
A particle starts with velocity \(\vec{u} = \left(3\hat{i} + 4\hat{j}\right)\) m/s and moves with a constant acceleration \(\vec{a} = \left(1\hat{i} - 2\hat{j}\right)\) m/s². Find (i) its velocity after 4 s, (ii) the magnitude of that velocity, and (iii) its displacement in these 4 s.
A swimmer can swim at 3 m/s in still water. He points himself due north across a river flowing due east at 4 m/s. Determine (i) his velocity relative to the ground in magnitude and direction, and (ii) the time he takes to cross the river if it is 60 m wide.
An aeroplane flies due east at 400 km/hr with respect to the air, while a wind blows from the south at 60 km/hr with respect to the ground. Determine the magnitude and direction of the velocity of the aeroplane with respect to the ground.
A ball is projected from level ground with a speed of 30 m/s at 60° to the horizontal. Taking \(g = 10\) m/s², find (i) the time of flight, (ii) the maximum height, (iii) the horizontal range, and (iv) the velocity of the ball 2 s after projection (magnitude and direction).
A stone is thrown horizontally with a speed of 12 m/s from the top of a cliff 45 m high. Taking \(g = 10\) m/s², find (i) the time it takes to reach the ground, (ii) the horizontal distance from the base of the cliff at which it lands, and (iii) its speed just before striking the ground.
A projectile launched at 40° to the horizontal has a horizontal range of 90 m. Taking \(g = 10\) m/s², find its speed of projection. What would the range have been for the same speed at 45°?
A particle moves along a circle of radius 1.5 m, completing 4 revolutions in 2 s with uniform speed. Find (i) its period, (ii) its angular speed, (iii) its linear speed, and (iv) the magnitude of its centripetal acceleration.
A stone of mass 0.2 kg tied to a string is whirled in a horizontal circle of radius 0.5 m with a uniform speed of 4 m/s. Calculate (i) the centripetal acceleration, and (ii) the tension in the string, assuming the string to be horizontal.
A car of mass 900 kg negotiates a level circular turn of radius 40 m at a uniform speed of 12 m/s. Calculate the centripetal force required. By what factor does this force change if the speed is doubled on the same turn?
A conical pendulum has a string of length 1.2 m, and the bob describes a horizontal circle with the string making an angle of 30° with the vertical. Taking \(g = 10\) m/s², determine (i) the radius of the circle, (ii) the vertical depth of the circle below the point of support, and (iii) the time period of revolution.
An aeroplane has a run of 500 m to take off from the runway. It starts from rest and moves with constant acceleration to cover the runway in 30 sec. What is the velocity of the aeroplane at take off?
A car moving along a straight road with a speed of 120 km/hr is brought to rest by applying brakes. The car covers a distance of 100 m before it stops. Calculate (i) the average retardation of the car, (ii) the time taken by the car to come to rest.
A car travels at a speed of 50 km/hr for 30 minutes, at 30 km/hr for the next 15 minutes, and then at 70 km/hr for the next 45 minutes. What is the average speed of the car?
A velocity-time graph is shown in the adjoining figure. Determine: (i) the initial speed of the car, (ii) the maximum speed attained by the car, (iii) the part of the graph showing zero acceleration, (iv) the part of the graph showing constant retardation, (v) the distance travelled by the car in the first 6 sec.
Diagram
Fig. 3.15 – Velocity-time graph for the car: v rises linearly from 0 at t = 0 to 20 m/s at t = 3 s (point A), stays constant at 20 m/s up to t = 6 s (point B), then falls linearly to 0 at t = 8 s (point C); points E and D marked on the time axis below A and B.
A man throws a ball to a maximum horizontal distance of 80 metres. Calculate the maximum height reached.
A particle is projected with speed \(v_0\) at an angle \(\theta\) to the horizontal on an inclined surface making an angle \(\phi\) (\(\phi < \theta\)) to the horizontal. Find the range of the projectile along the inclined surface.
A metro train runs from station A to B to C. It takes 4 minutes in travelling from station A to station B. The train halts at station B for 20 s. Then it starts from station B and reaches station C in the next 3 minutes. At the start, the train accelerates for 10 sec to reach the constant speed of 72 km/hr. The train moving at the constant speed is brought to rest in 10 sec at the next station. (i) Plot the velocity-time graph for the train travelling from station A to B to C. (ii) Calculate the distance between the stations A, B and C.
A train is moving eastward at 10 m/sec. A waiter is walking eastward at 1.2 m/sec, and a fly is flying toward the north across the waiter’s tray at 2 m/s. What is the velocity of the fly relative to the Earth?
A car moves in a circle at a constant speed of 50 m/s and completes one revolution in 40 s. Determine the magnitude of the acceleration of the car.
A particle moves in a circle with a constant speed of 15 m/s. The radius of the circle is 2 m. Determine the centripetal acceleration of the particle.
A projectile is thrown at an angle of 30° to the horizontal. What should be the range of initial velocity (u) so that its range will be between 40 m and 50 m? Assume \(g = 10\) m s⁻².
3.8 Objective Questions
- An object thrown from a moving bus is an example of
- Uniform circular motion
- Rectilinear motion
- Projectile motion
- Motion in one dimension
- For a particle having a uniform circular motion, which of the following is constant?
- Speed
- Acceleration
- Velocity
- Displacement
- The bob of a conical pendulum undergoes
- Rectilinear motion in a horizontal plane
- Uniform motion in a horizontal circle
- Uniform motion in a vertical circle
- Rectilinear motion in a vertical circle
- For uniform acceleration in rectilinear motion, which of the following is not correct?
- Velocity-time graph is linear
- Acceleration is the slope of the velocity-time graph
- The area under the velocity-time graph equals displacement
- Velocity-time graph is nonlinear
- If three particles A, B and C are having velocities \(\vec{v}_A\), \(\vec{v}_B\) and \(\vec{v}_C\), which of the following formulae gives the relative velocity of A with respect to B?
- \(\vec{v}_A + \vec{v}_B\)
- \(\vec{v}_A - \vec{v}_C + \vec{v}_B\)
- \(\vec{v}_A - \vec{v}_B\)
- \(\vec{v}_C - \vec{v}_A\)