3.1 Introduction
Chapter 3 — Motion in a Plane
On the evening of 7 August 2021, at the Tokyo Olympic Stadium, Neeraj Chopra ran down a narrow strip, planted his left foot, and hurled a spear-shaped javelin into the humid air. It rose, curved, and buried its metal tip in the grass at a distance of 87.58 metres — winning India its first-ever Olympic gold medal in athletics.
What decides how far a javelin flies? Not just how hard it is thrown. The direction of the throw matters just as much as the speed. A javelin launched straight up would rise high and land at the thrower’s own feet. One launched flat along the ground would skim into the turf almost at once. Somewhere between these two extremes lies the angle that carries it the farthest — and every javelin thrower spends years training their body to find it.
The physics of a javelin in flight is the physics of this chapter. To describe such motion, we can no longer use plus and minus signs on a single line, the way we did for motion along a straight road. We need a language that handles both magnitude and direction together. That language is the language of vectors. Once we learn to speak it, we can describe not only javelins but also raindrops slanting in the wind, cricket balls arcing to the boundary, satellites circling the Earth, and cars turning at a bend in the highway.
Figure to come
Fig. 3.0 – A silhouette of a javelin thrower at the moment of release, with the javelin’s parabolic path traced in the air toward a landing point on the ground; the launch angle from the horizontal is marked as θ₀ and the initial velocity as a vector arrow labelled v₀.
This chapter builds the tools to answer each of these questions.
In the previous chapter, we studied motion along a straight line. There we introduced four important quantities: position, displacement, velocity, and acceleration. Each of these has a direction, but because a straight line offers only two possible directions — forward or backward — we could handle the direction simply by placing a plus or a minus sign in front of the number.
That trick works only in one dimension. The moment an object begins to move in a plane, such as a football curving across a field or a boat crossing a river, “forward” and “backward” are no longer enough. The object could move north, south, east, west, or in any direction between them. A single sign cannot capture that.
To describe motion in two dimensions (a plane) or three dimensions (space), we therefore need a richer language — the language of vectors. A vector, unlike an ordinary number, carries both a size and a direction with it, and there are special rules for adding, subtracting, and multiplying vectors that do not look quite like ordinary arithmetic.
Our plan for this chapter is as follows. First, we learn what a vector is and how to combine vectors — how to add them, subtract them, and multiply them by ordinary numbers. What happens, for example, when you multiply a velocity vector by a duration of time? We shall see that the result is a displacement vector, and this kind of insight is exactly what makes vectors so useful.
Once we are comfortable with vectors, we return to the physics. We use vectors to define velocity and acceleration for an object moving in a plane. As the simplest interesting case, we study motion with a constant acceleration, and then apply this to projectile motion — the flight of any object thrown or launched into the air and moving under gravity alone.
Finally, we look at uniform circular motion — the motion of an object travelling at constant speed along a circle. Circular motion appears everywhere in daily life: the hands of a clock, the wheels of a car, a stone whirling on a string, satellites moving around the Earth. It has one surprising feature we will need to explain: an object can be moving at a perfectly steady speed and yet still be accelerating.
Everything we develop for motion in a plane extends naturally to motion in three-dimensional space. The vector language handles two and three dimensions equally well; only the number of components changes.
3.2 Scalars and Vectors
In physics, every measurable quantity falls into one of two families. Some quantities are fully described by a single number (with a unit) — they are called scalars. Others need a number and a direction to be fully described — they are called vectors. This one difference — the presence or absence of a direction — decides how we combine such quantities with each other, and it is the reason a plus/minus sign was enough in the last chapter but is not enough in this one.
A scalar quantity is a quantity with magnitude only. It is specified completely by a single number together with the proper unit. The distance between two points, the mass of an object, the temperature of a body, the time at which some event happens — all of these are scalars.
Because scalars have no direction, they combine using the ordinary rules of arithmetic. They can be added, subtracted, multiplied, and divided just like plain numbers — provided the units allow it.* For example, if the length and breadth of a rectangle are 1.0 m and 0.5 m, its perimeter is
\[1.0 \text{ m} + 0.5 \text{ m} + 1.0 \text{ m} + 0.5 \text{ m} = 3.0 \text{ m}\]
Each side is a scalar; the perimeter is a scalar too. Similarly, if the maximum and minimum temperatures on a certain day are 35.6 °C and 24.2 °C, the difference between them is 11.4 °C. And a uniform aluminium cube of side 10 cm with a mass of 2.7 kg has a volume of \(10^{-3}\text{ m}^3\) (a scalar) and a density of \(2.7 \times 10^3 \text{ kg m}^{-3}\) (a scalar).
Footnote: Addition and subtraction of scalars only make sense for quantities with the same unit. Multiplication and division, however, can be carried out across different units — that is how a distance divided by a time gives a speed.
Now consider quantities like displacement, velocity, acceleration, and force. Saying “the wind is blowing at 40 km/h” tells you only part of the story — from which direction? Saying “the plane is 300 km away” is incomplete — in which direction from where you are? These quantities carry a magnitude and a direction, and both pieces are needed to describe them.
The last part — “obeys the triangle law of addition” — is important. Not every quantity that seems to have a size and a direction qualifies as a vector; it must also combine in the special vector way we will study in the next few sections. This is why physicists distinguish carefully between quantities that only look like vectors and quantities that truly behave as vectors.
Notation. In this book, a vector is printed in bold face — for example, \(\mathbf{v}\) for a velocity vector. When writing by hand, bold face is hard to produce, so a small arrow is placed over the letter instead: \(\vec{v}\). Both \(\mathbf{v}\) and \(\vec{v}\) stand for the same quantity. The magnitude of a vector, sometimes called its absolute value, is written with vertical bars: \(|\mathbf{v}| = v\). So in the same passage you may see vectors written as \(\mathbf{A}, \mathbf{a}, \mathbf{p}, \mathbf{q}, \mathbf{r}, ..., \mathbf{x}, \mathbf{y}\), with their magnitudes as ordinary italic letters \(A, a, p, q, r, ..., x, y\).
3.2.1 Position and Displacement Vectors
To describe where an object is as it moves in a plane, we first fix a convenient reference point called the origin and label it O. Let P and P′ be the positions of the object at two instants, times \(t\) and \(t'\) respectively, as shown in Fig. 3.1(a). We draw a straight line from O to P and mark an arrow at its head. This directed line segment — from the origin O to the current position P — is called the position vector of the object at time \(t\). It is denoted by \(\mathbf{r}\), so that \(\mathbf{OP} = \mathbf{r}\). In the same way, the position vector of P′ at time \(t'\) is \(\mathbf{OP'} = \mathbf{r'}\).
Figure to come
Fig. 3.1(a) – A point O marked as origin with x and y axes drawn; a point P joined to O by an arrow labelled r; a nearby point P’ joined to O by an arrow labelled r’; a smooth curved arrow from P to P’ representing the object’s motion, with a straight arrow drawn from P to P’ labelled as the displacement.
The length of the position vector \(\mathbf{r}\) tells us how far the object is from the origin. The direction of \(\mathbf{r}\) tells us where P lies as seen from O.
Suppose the object now moves from P to P′. Draw an arrow with its tail at P and its tip at P′. This arrow is the displacement vector for the motion from P (at time \(t\)) to P′ (at time \(t'\)). It is written as \(\mathbf{PP'}\).
Notice the phrase “regardless of the actual path taken”. The displacement vector depends only on where the motion begins and where it ends — it does not care how the object got from one to the other. In Fig. 3.1(b), whether the object travels from P to Q along the winding path PABCQ, or the shorter PDQ, or the twisting PBEFQ, the displacement vector is the same straight arrow \(\mathbf{PQ}\) in every case.
Figure to come
Fig. 3.1(b) – Points P and Q joined by a straight arrow labelled PQ; three different curved paths from P to Q also shown lightly — one passing through points A, B, C, another through D, and another through B, E, F.
This has a useful consequence. Since a straight line is the shortest distance between two points, the magnitude of the displacement vector can never exceed the length of the path actually travelled. If the object moves in a straight line without turning back, the two are equal; in every other case, the path length is longer.
This same fact was noted in the previous chapter for motion in one dimension — it now holds for motion in a plane too, and for the same reason.
3.2.2 Equality of Vectors
Two vectors are equal only when both their magnitude and their direction match. Same length is not enough; same direction is not enough; both must agree.
Figure 3.2(a) shows two such vectors, \(\mathbf{A}\) and \(\mathbf{B}\). Their tails start at different points (O and Q), but their arrows are the same length and point the same way. To check the equality visually, slide \(\mathbf{B}\) parallel to itself — without rotating it — until its tail Q sits on top of O. If the tip of \(\mathbf{B}\) now sits exactly on the tip of \(\mathbf{A}\), the two vectors are indeed equal, and we write \(\mathbf{A} = \mathbf{B}\).
Figure to come
Fig. 3.2(a) – Two arrows of equal length pointing in the same direction; one with its tail at O and tip at P (labelled A), the other with its tail at Q and tip at S (labelled B).
The situation in Fig. 3.2(b) is different. Vectors \(\mathbf{A'}\) and \(\mathbf{B'}\) have equal lengths but point in noticeably different directions. Even if we slide \(\mathbf{B'}\) parallel to itself so that its tail Q′ meets O′, the tip S′ of \(\mathbf{B'}\) will not land on the tip P′ of \(\mathbf{A'}\). So \(\mathbf{A'} \neq \mathbf{B'}\), even though \(|\mathbf{A'}| = |\mathbf{B'}|\).
Figure to come
Fig. 3.2(b) – Two arrows of equal length but pointing in clearly different directions; one with tail O’ and tip P’ (labelled A’), the other with tail Q’ and tip S’ (labelled B’).
The sliding trick works only because, in our study, a vector is not tied to any particular location in space. You are free to translate it (that is, move it without rotating it) and the vector itself is unchanged. Vectors of this kind are called free vectors. In some advanced applications — such as analysing forces that act along specific lines on a rigid body — the exact line along which a vector acts also matters, and such vectors are called localised vectors. Unless stated otherwise, every vector in this chapter is a free vector.
3.3 Multiplication of Vectors by Real Numbers
Once we have vectors, a natural question arises: what happens when we multiply a vector by an ordinary number? Does the result stay a vector, and if so, how do its magnitude and direction change?
Multiplication of a vector by a real number turns out to stretch, shrink, or flip the vector — but never otherwise rotates it. The direction is either preserved or reversed; the magnitude simply scales.
Consider a vector \(\mathbf{A}\) and a positive real number \(\lambda\). Multiplying \(\mathbf{A}\) by \(\lambda\) gives a new vector \(\lambda\mathbf{A}\) whose magnitude is \(\lambda\) times that of \(\mathbf{A}\), and whose direction is exactly the same as \(\mathbf{A}\):
\[|\lambda \mathbf{A}| = \lambda\, |\mathbf{A}| \quad \text{if } \lambda > 0.\]
For example, multiplying \(\mathbf{A}\) by 2 gives \(2\mathbf{A}\), which points the same way as \(\mathbf{A}\) and is twice as long, as shown in Fig. 3.3(a).
Figure to come
Fig. 3.3(a) – A horizontal arrow labelled A pointing to the right, and below it a second arrow of twice its length labelled 2A, pointing the same way.
If the multiplier is negative, the direction reverses. Multiplying \(\mathbf{A}\) by \(-\lambda\) (with \(\lambda > 0\)) gives a vector of magnitude \(\lambda\,|\mathbf{A}|\) but pointing opposite to \(\mathbf{A}\). In particular, \(-\mathbf{A}\) has the same length as \(\mathbf{A}\) but points the opposite way, and \(-1.5\,\mathbf{A}\) is 1.5 times as long, again reversed. Both are shown in Fig. 3.3(b).
Figure to come
Fig. 3.3(b) – Vector A pointing to the right, and below it vectors −A (same length as A, pointing left) and −1.5A (1.5 times the length of A, also pointing left).
The multiplier \(\lambda\) need not be a pure number — it can itself be a physical quantity with its own dimensions. In that case, the dimensions of the resulting vector \(\lambda\mathbf{A}\) are the product of the dimensions of \(\lambda\) and the dimensions of \(\mathbf{A}\).
For instance, suppose a car moves at a constant velocity \(\mathbf{v}\) (dimensions of length/time). If the motion continues for a duration \(t\) (dimensions of time), then \(\mathbf{v}\,t\) has dimensions of length — the dimension of displacement. And indeed, \(\mathbf{v}\,t\) is the displacement of the car during that time. The direction of \(\mathbf{v}\) is preserved, but the physical meaning of the resulting vector has changed from a velocity into a displacement.
3.4 Addition and Subtraction of Vectors — Graphical Method
We already know from Section 3.2 that vectors, by definition, obey the triangle law of addition (or equivalently, the parallelogram law). It is now time to see what these laws actually say geometrically and how to use them in practice.
Suppose we have two vectors \(\mathbf{A}\) and \(\mathbf{B}\) lying in a plane, as shown in Fig. 3.4(a). The lengths of the arrows are drawn proportional to the magnitudes of the two vectors.
Figure to come
Fig. 3.4(a) – Two arrows A and B drawn separately in a plane, not touching each other, with their tails at different points.
The head-to-tail (triangle) method
To find the sum \(\mathbf{A} + \mathbf{B}\), do the following: without rotating \(\mathbf{B}\), slide it so that its tail sits at the head of \(\mathbf{A}\), as in Fig. 3.4(b). Now draw a new arrow starting from the tail of \(\mathbf{A}\) and ending at the head of the just-placed \(\mathbf{B}\). This closing arrow is the vector sum:
\[\mathbf{R} = \mathbf{A} + \mathbf{B}\]
Because the vectors are arranged head to tail, this construction is called the head-to-tail method. And because the two original vectors and the resultant form the three sides of a triangle, it is also called the triangle method of vector addition.
Figure to come
Fig. 3.4(b) – Vector A drawn from O to P; vector B drawn starting at P and ending at Q. A diagonal arrow from O to Q, labelled R = A + B, closes the triangle.
Now what if we add the vectors in the reverse order, \(\mathbf{B} + \mathbf{A}\)? Place \(\mathbf{B}\) first and then attach \(\mathbf{A}\) to its head, as shown in Fig. 3.4(c). The closing arrow — again from the tail of the first to the head of the second — has the same length and direction as before. So the answer is the same:
\[\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A} \quad (3.1)\]
Vector addition is thus commutative — the order in which you add does not change the result.
Figure to come
Fig. 3.4(c) – Vector B drawn from S to Q, then vector A from Q to P. Diagonal arrow from S to P shown, equal to the earlier resultant R.
Vector addition also obeys the associative law. Given three vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\), adding \(\mathbf{A}\) and \(\mathbf{B}\) first and then adding \(\mathbf{C}\) gives the same result as adding \(\mathbf{B}\) and \(\mathbf{C}\) first and then adding \(\mathbf{A}\):
\[(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C}) \quad (3.2)\]
This is illustrated in Fig. 3.4(d).
Figure to come
Fig. 3.4(d) – Three vectors A, B, C connected head-to-tail as a broken path, with the intermediate sums (A+B) and (B+C) drawn as diagonals, and the total resultant A+B+C drawn once, showing that both bracketings give the same arrow.
The null (zero) vector
What happens when we add two vectors of equal magnitude but opposite direction? Take a vector \(\mathbf{A}\) and its opposite \(-\mathbf{A}\), shown earlier in Fig. 3.3(b). Their sum is \(\mathbf{A} + (-\mathbf{A})\). Since the two arrows have the same length but point in opposite directions, placing them head to tail brings you right back to where you started. The resultant has zero magnitude.
This zero-magnitude vector is called the null vector or zero vector, and is written as \(\mathbf{0}\):
\[\mathbf{A} - \mathbf{A} = \mathbf{0}, \qquad |\mathbf{0}| = 0 \quad (3.3)\]
A null vector also appears whenever we multiply any vector \(\mathbf{A}\) by the scalar 0. The main properties of \(\mathbf{0}\) are:
\[\mathbf{A} + \mathbf{0} = \mathbf{A}\] \[\lambda\, \mathbf{0} = \mathbf{0}\] \[0\, \mathbf{A} = \mathbf{0} \quad (3.4)\]
What does a null vector mean physically? Look again at Fig. 3.1(a). Suppose an object at P at time \(t\) moves to P′ and then comes all the way back to P. What is its total displacement? Since the initial and final positions coincide, the displacement vector has zero magnitude — that is, it is a null vector. The object may have covered a large path length, but its net displacement is zero.
Subtraction of vectors
Subtracting one vector from another is defined simply as adding the negative of the second vector:
\[\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B}) \quad (3.5)\]
To do this graphically, first construct \(-\mathbf{B}\) by reversing the direction of \(\mathbf{B}\). Then add it to \(\mathbf{A}\) using the head-to-tail rule. Fig. 3.5 shows both the difference \(\mathbf{R_2} = \mathbf{A} - \mathbf{B}\) and, for comparison, the sum \(\mathbf{R_1} = \mathbf{A} + \mathbf{B}\) drawn in the same figure.
Figure to come
Fig. 3.5 – Vector A pointing to the upper right, vector B pointing downward, and vector −B (opposite of B) also shown. Two resultants are drawn: R₁ = A + B (from tail of A to head of B) and R₂ = A − B (from tail of A to head of −B).
The parallelogram method
There is a second, equally common way to add two vectors, called the parallelogram method. Instead of placing one vector at the tail of the other, we bring the tails of both vectors to a common origin O, as shown in Fig. 3.6(a).
Figure to come
Fig. 3.6(a) – Vector A along OP and vector B along OQ, both drawn with their tails at the common origin O.
Now, from the head of \(\mathbf{A}\), draw a line parallel to \(\mathbf{B}\); from the head of \(\mathbf{B}\), draw a line parallel to \(\mathbf{A}\). These two lines meet at some point S, completing a parallelogram OQSP. Join the origin O to the far corner S. The arrow \(\mathbf{OS}\) is the resultant \(\mathbf{R}\), as shown in Fig. 3.6(b):
\[\mathbf{R} = \mathbf{A} + \mathbf{B}\]
Figure to come
Fig. 3.6(b) – Same setup with the parallelogram OQSP completed by two dashed lines; the diagonal from O to S, labelled R, is drawn as a bold arrow.
To see that this gives the same answer as the triangle method, look at the triangle OQS in Fig. 3.6(c). The side QS of the parallelogram is parallel to OP and of the same length, so QS is a copy of the vector \(\mathbf{A}\). In the triangle OQS, then, we have \(\mathbf{B}\) from O to Q, followed by \(\mathbf{A}\) from Q to S, arranged head-to-tail. Their sum by the triangle method is exactly \(\mathbf{OS}\) — which is the same diagonal we obtained from the parallelogram. So the two methods are equivalent.
Figure to come
Fig. 3.6(c) – Same parallelogram OQSP with vectors B (from O to Q) and A (from Q to S) now emphasized as a triangle OQS, showing that OS is the resultant, matching the parallelogram diagonal.
3.5 Resolution of Vectors
The last few sections showed how two vectors can be added to give a single resultant. Now consider the opposite question. Given a single vector, can we always break it down into two (or more) pieces along directions of our choice? The answer is yes, and the process is called resolution — literally, resolving one vector into components.
Resolution is not just a mathematical trick. In physics, we constantly need to analyse a single physical vector — a force, a velocity, the weight of an object on a slope — along directions that are convenient for the problem. Resolving a vector into components makes the algebra of physics dramatically simpler.
Resolution along two arbitrary directions
Let \(\mathbf{a}\) and \(\mathbf{b}\) be any two non-zero vectors in a plane, pointing in different directions. Let \(\mathbf{A}\) be any other vector in the same plane, as shown in Fig. 3.8(a). We shall show that \(\mathbf{A}\) can always be written as a sum of two vectors — one parallel to \(\mathbf{a}\) and one parallel to \(\mathbf{b}\).
Figure to come
Fig. 3.8(a) – Two non-collinear vectors a and b drawn from a common origin, pointing in different directions.
Let O and P be the tail and the head of \(\mathbf{A}\). Through O, draw a straight line parallel to \(\mathbf{a}\); through P, draw a straight line parallel to \(\mathbf{b}\). These two lines meet at some point Q, as shown in Fig. 3.8(b). By the triangle law of vector addition,
\[\mathbf{A} = \mathbf{OP} = \mathbf{OQ} + \mathbf{QP} \quad (3.6)\]
Figure to come
Fig. 3.8(b) – Vector A drawn from O to P. Through O, a line parallel to a extended to Q; through P, a line parallel to b meeting the first line at Q. The segments OQ and QP are highlighted as λa and μb.
Now, \(\mathbf{OQ}\) is parallel to \(\mathbf{a}\), so it must be some scalar multiple of \(\mathbf{a}\). Similarly, \(\mathbf{QP}\) is parallel to \(\mathbf{b}\), so it must be some scalar multiple of \(\mathbf{b}\). Writing these multiples as \(\lambda\) and \(\mu\),
\[\mathbf{OQ} = \lambda \mathbf{a}, \qquad \mathbf{QP} = \mu \mathbf{b} \quad (3.7)\]
Substituting into Eq. (3.6),
\[\mathbf{A} = \lambda \mathbf{a} + \mu \mathbf{b} \quad (3.8)\]
We say that \(\mathbf{A}\) has been resolved into two component vectors \(\lambda \mathbf{a}\) and \(\mu \mathbf{b}\) along the directions of \(\mathbf{a}\) and \(\mathbf{b}\) respectively. The three vectors \(\mathbf{A}\), \(\mathbf{a}\), and \(\mathbf{b}\) all lie in the same plane; the values of \(\lambda\) and \(\mu\) are uniquely determined once the directions of \(\mathbf{a}\) and \(\mathbf{b}\) are fixed.
Unit vectors
While Eq. (3.8) works for any pair of non-collinear directions, in practice it is far more convenient to resolve a vector along the axes of a rectangular coordinate system. For that, we need special vectors that carry only direction information, with no magnitude of their own to worry about. These are called unit vectors.
The unit vectors along the x-, y-, and z-axes of a rectangular coordinate system are denoted by \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), and \(\hat{\mathbf{k}}\) respectively, as shown in Fig. 3.9(a). Since each has magnitude 1,
\[|\hat{\mathbf{i}}| = |\hat{\mathbf{j}}| = |\hat{\mathbf{k}}| = 1 \quad (3.9)\]
Figure to come
Fig. 3.9(a) – A right-handed rectangular coordinate system with the three unit vectors î along +x, ĵ along +y, and k̂ along +z.
These three unit vectors are mutually perpendicular. In print, they are shown in bold face with a cap (^) to distinguish them from ordinary vectors. Since this chapter deals with motion in two dimensions, we shall mostly need only two of them — \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\).
Multiplying any unit vector \(\hat{\mathbf{n}}\) by a scalar \(\lambda\) produces a vector of magnitude \(\lambda\) pointing along \(\hat{\mathbf{n}}\). In particular, any vector \(\mathbf{A}\) can be written as
\[\mathbf{A} = |\mathbf{A}|\,\hat{\mathbf{n}} \quad (3.10)\]
where \(\hat{\mathbf{n}}\) is the unit vector in the direction of \(\mathbf{A}\). Every vector, in other words, is a magnitude times a direction — with the unit vector carrying the direction and \(|\mathbf{A}|\) carrying the magnitude.
Resolution along the x- and y-axes
Consider a vector \(\mathbf{A}\) lying in the x-y plane, as shown in Fig. 3.9(b). From the head of \(\mathbf{A}\), drop perpendiculars onto the x-axis and the y-axis. This produces two vectors \(\mathbf{A_1}\) and \(\mathbf{A_2}\) — one along the x-axis, the other along the y-axis — such that
\[\mathbf{A_1} + \mathbf{A_2} = \mathbf{A}\]
Figure to come
Fig. 3.9(b) – Vector A starting at origin O in the x-y plane, with perpendiculars dropped from its tip to the x- and y-axes forming a rectangle. The projections along the axes are shown as A₁ (of length A₁ along x) and A₂ (of length A₂ along y).
Since \(\mathbf{A_1}\) is parallel to \(\hat{\mathbf{i}}\) and \(\mathbf{A_2}\) is parallel to \(\hat{\mathbf{j}}\), we can write
\[\mathbf{A_1} = A_x \hat{\mathbf{i}}, \qquad \mathbf{A_2} = A_y \hat{\mathbf{j}} \quad (3.11)\]
where \(A_x\) and \(A_y\) are real numbers. Therefore,
\[\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} \quad (3.12)\]
This is shown in Fig. 3.9(c). The quantities \(A_x\) and \(A_y\) are called the x-component and y-component of the vector \(\mathbf{A}\).
Figure to come
Fig. 3.9(c) – Same as Fig. 3.9(b) but with vectors labelled A_x î along the x-axis and A_y ĵ along the y-axis. The angle θ between A and the positive x-axis is marked.
A subtle point: \(A_x\) by itself is a number (positive, negative, or zero) — it is not a vector. But \(A_x \hat{\mathbf{i}}\) is a vector, because it carries a direction through \(\hat{\mathbf{i}}\). Similarly for \(A_y \hat{\mathbf{j}}\).
Using simple trigonometry on the right-angled triangle formed by \(\mathbf{A}\), \(\mathbf{A_1}\), and \(\mathbf{A_2}\), we can express \(A_x\) and \(A_y\) in terms of the magnitude \(A\) and the angle \(\theta\) that \(\mathbf{A}\) makes with the positive x-axis:
\[A_x = A\cos\theta, \qquad A_y = A\sin\theta \quad (3.13)\]
Because \(\sin\theta\) and \(\cos\theta\) can each be positive, negative, or zero depending on \(\theta\), so can the components \(A_x\) and \(A_y\). For example, a vector pointing into the third quadrant (angle between 180° and 270°) has both components negative.
Two ways to specify a vector in a plane
Equations (3.12) and (3.13) reveal an important fact: a vector in a plane can be described completely in either of two equivalent ways.
- By its magnitude and direction: give \(A\) and the angle \(\theta\) it makes with the x-axis.
- By its components: give \(A_x\) and \(A_y\).
Each description determines the other. If \(A\) and \(\theta\) are known, Eq. (3.13) gives \(A_x\) and \(A_y\) directly. Going the other way, from components to magnitude and direction, we use Pythagoras’ theorem and inverse tangent. Squaring and adding the two equations in Eq. (3.13),
\[A_x^2 + A_y^2 = A^2\cos^2\theta + A^2\sin^2\theta = A^2\]
so that
\[A = \sqrt{A_x^2 + A_y^2} \quad (3.14)\]
And
\[\tan\theta = \frac{A_y}{A_x}, \qquad \theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right) \quad (3.15)\]
Together, Eqs. (3.14) and (3.15) let us recover magnitude and direction from the components. In this chapter, we shall move freely between the two descriptions, choosing whichever is more convenient for the problem at hand.
Resolution in three dimensions
So far we have considered vectors lying in a plane. The same procedure carries over to three dimensions, with just one more axis. If \(\alpha\), \(\beta\), and \(\gamma\) are the angles* that a vector \(\mathbf{A}\) makes with the x-, y-, and z-axes respectively, as shown in Fig. 3.9(d), then its three components are
\[A_x = A\cos\alpha, \qquad A_y = A\cos\beta, \qquad A_z = A\cos\gamma \quad (3.16a)\]
Figure to come
Fig. 3.9(d) – Vector A drawn from the origin inside a 3D coordinate box, with dashed lines dropped from its tip to the three axes. Angles α, β, γ between A and the x-, y-, and z-axes are marked, and the projections A_x, A_y, A_z are shown.
The vector itself is written as
\[\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}} \quad (3.16b)\]
and its magnitude is given by the three-dimensional form of Pythagoras’ theorem,
\[A = \sqrt{A_x^2 + A_y^2 + A_z^2} \quad (3.16c)\]
Footnote: The angles \(\alpha\), \(\beta\), and \(\gamma\) are three-dimensional angles — that is, angles between the vector and each of the axes taken as pairs of lines in space, which are not coplanar.
A particularly useful application of Eq. (3.16b) is the position vector of a point in space. If a point has coordinates \((x, y, z)\) measured from the origin, then its position vector is
\[\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}} \quad (3.17)\]
where \(x\), \(y\), and \(z\) are the components of \(\mathbf{r}\) along the three coordinate axes. Every point in space, then, corresponds to a unique triple of numbers, and vice versa.
3.6 Vector Addition – Analytical Method
The graphical methods of Section 3.4 — head-to-tail and parallelogram — are excellent for building intuition. But they have two practical drawbacks. First, they depend on drawing arrows to scale, so the accuracy of the result is limited by the ruler and the protractor. Second, when three, four, or more vectors need to be added, the graphical construction quickly becomes tedious.
There is a far more efficient way that uses the resolution of vectors into rectangular components (Section 3.5). We simply add the components of the vectors, axis by axis. This is called the analytical method of vector addition.
Adding two vectors in a plane
Consider two vectors \(\mathbf{A}\) and \(\mathbf{B}\) in the x-y plane, with components \((A_x, A_y)\) and \((B_x, B_y)\):
\[\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} \quad (3.18)\] \[\mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}}\]
Let \(\mathbf{R}\) be their sum. Then
\[\mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}}) + (B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}}) \quad (3.19a)\]
Because vector addition is both commutative and associative (Section 3.4), we can regroup the terms freely — gathering all the \(\hat{\mathbf{i}}\) pieces together and all the \(\hat{\mathbf{j}}\) pieces together:
\[\mathbf{R} = (A_x + B_x)\hat{\mathbf{i}} + (A_y + B_y)\hat{\mathbf{j}} \quad (3.19b)\]
But we already know that \(\mathbf{R}\) itself can be written as
\[\mathbf{R} = R_x \hat{\mathbf{i}} + R_y \hat{\mathbf{j}} \quad (3.20)\]
Comparing Eqs. (3.19b) and (3.20), we get the simple result
\[R_x = A_x + B_x, \qquad R_y = A_y + B_y \quad (3.21)\]
In words: each component of the resultant vector is just the sum of the corresponding components of the vectors being added. No geometry, no scale drawing — just ordinary addition, done separately along each axis.
Extension to three dimensions
The same idea works in three dimensions. If
\[\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}\] \[\mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}}\]
then
\[\mathbf{R} = \mathbf{A} + \mathbf{B} = R_x \hat{\mathbf{i}} + R_y \hat{\mathbf{j}} + R_z \hat{\mathbf{k}}\]
with
\[R_x = A_x + B_x, \qquad R_y = A_y + B_y, \qquad R_z = A_z + B_z \quad (3.22)\]
Once again, componentwise addition — just three additions instead of two.
More than two vectors, and subtraction
Nothing about the argument depended on there being only two vectors. Any number of vectors can be added this way, and subtraction is handled by simply putting a minus sign on the components of the vector to be subtracted.
For instance, if three vectors are given as
\[\mathbf{a} = a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}} + a_z \hat{\mathbf{k}}\] \[\mathbf{b} = b_x \hat{\mathbf{i}} + b_y \hat{\mathbf{j}} + b_z \hat{\mathbf{k}}\] \[\mathbf{c} = c_x \hat{\mathbf{i}} + c_y \hat{\mathbf{j}} + c_z \hat{\mathbf{k}} \quad (3.23a)\]
then the vector \(\mathbf{T} = \mathbf{a} + \mathbf{b} - \mathbf{c}\) has components
\[T_x = a_x + b_x - c_x\] \[T_y = a_y + b_y - c_y \quad (3.23b)\] \[T_z = a_z + b_z - c_z\]
Every component is handled independently. This is why the analytical method is the workhorse of every serious physics computation involving vectors.
3.7 Motion in a Plane
We now have all the vector tools we need. In this section, we use those tools to describe motion in a plane — that is, motion of an object whose position can change in two dimensions at once.
3.7.1 Position Vector and Displacement
To describe where a particle is at any moment, we place an origin, set up an x-y reference frame, and draw the position vector from the origin to the particle.
For a particle P located in the x-y plane, the position vector \(\mathbf{r}\) is (Fig. 3.12(a))
\[\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}\]
where \(x\) and \(y\) are the components of \(\mathbf{r}\) along the x- and y-axes. They are the same as the coordinates of the particle.
Figure to come
Fig. 3.12(a) – Origin O with x and y axes; a point P shown in the first quadrant; the position vector r drawn from O to P; horizontal projection xî and vertical projection yĵ shown.
Now suppose the particle moves along a curved path in the plane, as shown by the thick line in Fig. 3.12(b). Let it be at P at time \(t\) and at P′ at a later time \(t'\). The displacement of the particle during this interval is the straight vector from P to P′, denoted \(\Delta \mathbf{r}\):
\[\Delta \mathbf{r} = \mathbf{r'} - \mathbf{r} \quad (3.25)\]
The arrow points from P towards P′.
Figure to come
Fig. 3.12(b) – Origin O with axes; a curved path in the plane; position vectors r and r’ drawn from O to points P and P′ on the path; the straight vector Δr drawn from P to P′; components Δx and Δy shown; an arrow at the tail of Δr labelled “direction of v̄”.
In component form, using \(\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}\) and \(\mathbf{r'} = x'\hat{\mathbf{i}} + y'\hat{\mathbf{j}}\),
\[\Delta \mathbf{r} = (x'\hat{\mathbf{i}} + y'\hat{\mathbf{j}}) - (x\hat{\mathbf{i}} + y\hat{\mathbf{j}}) = \hat{\mathbf{i}}\,\Delta x + \hat{\mathbf{j}}\,\Delta y\]
where
\[\Delta x = x' - x, \qquad \Delta y = y' - y \quad (3.26)\]
Notice that the displacement depends only on the initial and final positions, not on the actual curved path taken between them — a fact we already saw in Section 3.2.1.
Velocity
Average velocity. The average velocity \(\overline{\mathbf{v}}\) of an object is the ratio of the displacement to the corresponding time interval:
\[\overline{\mathbf{v}} = \frac{\Delta \mathbf{r}}{\Delta t} = \frac{\Delta x\,\hat{\mathbf{i}} + \Delta y\,\hat{\mathbf{j}}}{\Delta t} = \hat{\mathbf{i}}\,\frac{\Delta x}{\Delta t} + \hat{\mathbf{j}}\,\frac{\Delta y}{\Delta t} \quad (3.27)\]
Or, writing it in components,
\[\overline{\mathbf{v}} = \overline{v}_x\hat{\mathbf{i}} + \overline{v}_y\hat{\mathbf{j}}\]
Since \(\Delta t\) is a positive scalar, dividing \(\Delta \mathbf{r}\) by \(\Delta t\) does not change its direction — only its magnitude. So the average velocity vector points in the same direction as the displacement vector \(\Delta \mathbf{r}\) (Fig. 3.12(b)).
Instantaneous velocity. The average velocity is not always what we want. If the particle’s speed or direction is changing during \(\Delta t\), then \(\overline{\mathbf{v}}\) is only an average over the whole interval — it does not tell us how fast, or in what direction, the particle is moving at a particular instant.
To get the velocity at an instant, we shrink \(\Delta t\) to zero:
\[\mathbf{v} = \lim_{\Delta t \to 0}\frac{\Delta \mathbf{r}}{\Delta t} = \frac{d\mathbf{r}}{dt} \quad (3.28)\]
This is called the (instantaneous) velocity.
Direction of the velocity is along the tangent. The limiting process behind Eq. (3.28) is worth pausing over. Look at Figs. 3.13(a) to (d). In each panel, the thick line is the actual path of the particle. Positions P, P₁, P₂, P₃ correspond to times \(t\), \(t + \Delta t_1\), \(t + \Delta t_2\), \(t + \Delta t_3\), where \(\Delta t_1 > \Delta t_2 > \Delta t_3\). The average velocity \(\overline{\mathbf{v}}\) over each interval points along the corresponding chord PP₁, PP₂, PP₃ — always along the straight arrow from the starting point to the finishing point.
Figure to come
Fig. 3.13(a), (b), (c) – Same curved path shown three times, with points P and P_i on the curve; the chord from P to each P_i drawn as Δr, with the direction of average velocity along it; the chord comes closer to being a tangent as ∆t decreases. Panel (d) shows the limit: v drawn as a tangent to the path at P.
As \(\Delta t\) becomes smaller and smaller, the second point P′ slides towards P along the curve, and the chord — which was the straight line joining them — becomes closer and closer to the tangent line at P. In the limit, it is the tangent. Therefore:
Component form. Expanding Eq. (3.28) in components,
\[\mathbf{v} = \frac{d\mathbf{r}}{dt} = \lim_{\Delta t \to 0}\!\left(\frac{\Delta x}{\Delta t}\hat{\mathbf{i}} + \frac{\Delta y}{\Delta t}\hat{\mathbf{j}}\right) \quad (3.29)\]
\[= \hat{\mathbf{i}}\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} + \hat{\mathbf{j}}\lim_{\Delta t \to 0}\frac{\Delta y}{\Delta t}\]
Or,
\[\mathbf{v} = \hat{\mathbf{i}}\frac{dx}{dt} + \hat{\mathbf{j}}\frac{dy}{dt} = v_x\hat{\mathbf{i}} + v_y\hat{\mathbf{j}}\]
where
\[v_x = \frac{dx}{dt}, \qquad v_y = \frac{dy}{dt} \quad (3.30a)\]
So if the position of the particle is given as a function of time — that is, if we know \(x(t)\) and \(y(t)\) — then differentiating gives the two components of the velocity directly.
The magnitude of \(\mathbf{v}\) is
\[v = \sqrt{v_x^2 + v_y^2} \quad (3.30b)\]
and the direction of \(\mathbf{v}\) makes an angle \(\theta\) with the x-axis given by
\[\tan\theta = \frac{v_y}{v_x}, \qquad \theta = \tan^{-1}\!\left(\frac{v_y}{v_x}\right) \quad (3.30c)\]
\(v_x\), \(v_y\), and \(\theta\) are shown in Fig. 3.14 for a velocity vector \(\mathbf{v}\) at a point \(\mathbf{p}\). These satisfy \(v_x = v\cos\theta\) and \(v_y = v\sin\theta\), just like the components of any vector.
Figure to come
Fig. 3.14 – A curved path with a point P on it. A tangent arrow v drawn at P, with its components v_x î (horizontal) and v_y ĵ (vertical) shown; the angle θ between v and the x-axis marked.
Acceleration
Average acceleration. When the velocity of a particle in a plane changes — either in magnitude or in direction — the particle is said to be accelerating. The average acceleration \(\overline{\mathbf{a}}\) for a time interval \(\Delta t\) is
\[\overline{\mathbf{a}} = \frac{\Delta \mathbf{v}}{\Delta t} = \frac{\Delta(v_x\hat{\mathbf{i}} + v_y\hat{\mathbf{j}})}{\Delta t} = \frac{\Delta v_x}{\Delta t}\hat{\mathbf{i}} + \frac{\Delta v_y}{\Delta t}\hat{\mathbf{j}} \quad (3.31a)\]
Or,
\[\overline{\mathbf{a}} = \overline{a}_x\hat{\mathbf{i}} + \overline{a}_y\hat{\mathbf{j}} \quad (3.31b)\]
Instantaneous acceleration. As with velocity, the instantaneous acceleration is obtained by taking the limit \(\Delta t \to 0\):
\[\mathbf{a} = \lim_{\Delta t \to 0}\frac{\Delta \mathbf{v}}{\Delta t} \quad (3.32a)\]
Since \(\Delta \mathbf{v} = \Delta v_x\,\hat{\mathbf{i}} + \Delta v_y\,\hat{\mathbf{j}}\),
\[\mathbf{a} = \hat{\mathbf{i}}\lim_{\Delta t \to 0}\frac{\Delta v_x}{\Delta t} + \hat{\mathbf{j}}\lim_{\Delta t \to 0}\frac{\Delta v_y}{\Delta t}\]
Or,
\[\mathbf{a} = a_x\hat{\mathbf{i}} + a_y\hat{\mathbf{j}} \quad (3.32b)\]
where
\[a_x = \frac{dv_x}{dt}, \qquad a_y = \frac{dv_y}{dt} \quad (3.32c)\]
Footnote: In terms of the coordinates \(x\) and \(y\), the components of acceleration can also be written as second derivatives of position:
\[a_x = \frac{d}{dt}\!\left(\frac{dx}{dt}\right) = \frac{d^2x}{dt^2}, \qquad a_y = \frac{d}{dt}\!\left(\frac{dy}{dt}\right) = \frac{d^2y}{dt^2}\]
Direction of the acceleration. Just as with velocity, we can understand the limiting process for acceleration graphically. In Figs. 3.15(a) to (d), P is the position at time \(t\) and P₁, P₂, P₃ are positions after \(\Delta t_1\), \(\Delta t_2\), \(\Delta t_3\) (with \(\Delta t_1 > \Delta t_2 > \Delta t_3\)). The velocity vectors at P, P₁, P₂, P₃ are drawn. For each interval, the change in velocity \(\Delta \mathbf{v}\) is found by the triangle law of vector subtraction (Section 3.4), and the direction of the average acceleration is along that \(\Delta \mathbf{v}\).
Figure to come
Fig. 3.15(a), (b), (c) – The curved path with velocity vectors v and v’ drawn at P and each P_i; the change Δv obtained by triangle-law subtraction; as ∆t decreases, the direction of Δv rotates. Panel (d) shows the limit: instantaneous acceleration a drawn as a vector at P.
As \(\Delta t\) shrinks, the direction of \(\Delta \mathbf{v}\) typically rotates, and in the limit \(\Delta t \to 0\) [Fig. 3.15(d)] the average acceleration becomes the instantaneous acceleration \(\mathbf{a}\) at the point P. Its direction depends entirely on how the velocity is changing at that instant — and, unlike the velocity itself, the acceleration is not in general along the tangent to the path.
This is a big difference from one-dimensional motion. In 1D, the velocity and the acceleration lie along the same straight line — they are either parallel (same direction) or anti-parallel (opposite direction). But for motion in two or three dimensions, \(\mathbf{v}\) and \(\mathbf{a}\) can point in any two directions, making any angle between \(0°\) and \(180°\) with each other. We shall see this vividly in projectile motion (Section 3.9), where the velocity constantly changes direction while the acceleration points straight down throughout the flight.
3.8 Motion in a Plane with Constant Acceleration
In Section 3.7 we defined velocity and acceleration in a plane in full generality. We now specialize to a particularly important case: motion in which the acceleration is constant — same magnitude, same direction, throughout the motion. This is the two-dimensional analogue of the uniformly accelerated motion studied in the previous chapter.
Velocity as a function of time
Suppose an object moves in the x-y plane and its acceleration \(\mathbf{a}\) is constant. Over any time interval, the average acceleration equals this constant value. Let the velocity be \(\mathbf{v_0}\) at time \(t = 0\) and \(\mathbf{v}\) at time \(t\). Then, by the definition of acceleration,
\[\mathbf{a} = \frac{\mathbf{v} - \mathbf{v_0}}{t - 0} = \frac{\mathbf{v} - \mathbf{v_0}}{t}\]
Rearranging,
\[\mathbf{v} = \mathbf{v_0} + \mathbf{a}\, t \quad (3.33a)\]
This is the vector form of the first kinematic equation. In terms of components,
\[v_x = v_{0x} + a_x\, t\] \[v_y = v_{0y} + a_y\, t \quad (3.33b)\]
Position as a function of time
Next, we find how the position vector \(\mathbf{r}\) changes with time. The idea is the same as in one dimension. Let \(\mathbf{r_0}\) and \(\mathbf{r}\) be the position vectors at times 0 and \(t\), and let the velocities at these instants be \(\mathbf{v_0}\) and \(\mathbf{v}\). When the acceleration is constant, the average velocity over the interval equals the arithmetic mean of the initial and final velocities:
\[\overline{\mathbf{v}} = \frac{\mathbf{v_0} + \mathbf{v}}{2}\]
The displacement is this average velocity multiplied by the time interval:
\[\mathbf{r} - \mathbf{r_0} = \frac{\mathbf{v_0} + \mathbf{v}}{2}\, t = \frac{\mathbf{v_0} + (\mathbf{v_0} + \mathbf{a}\, t)}{2}\, t = \mathbf{v_0}\, t + \frac{1}{2}\,\mathbf{a}\, t^2\]
Therefore,
\[\mathbf{r} = \mathbf{r_0} + \mathbf{v_0}\, t + \frac{1}{2}\,\mathbf{a}\, t^2 \quad (3.34a)\]
This is the vector form of the second kinematic equation. It can be easily verified: differentiating Eq. (3.34a) with respect to \(t\) gives \(d\mathbf{r}/dt = \mathbf{v_0} + \mathbf{a}\, t = \mathbf{v}\), which is Eq. (3.33a). Also, at \(t = 0\) the equation gives \(\mathbf{r} = \mathbf{r_0}\), as expected.
In component form,
\[x = x_0 + v_{0x}\, t + \frac{1}{2}\, a_x\, t^2\] \[y = y_0 + v_{0y}\, t + \frac{1}{2}\, a_y\, t^2 \quad (3.34b)\]
The key insight: independence of the x- and y-motions
Look at Eq. (3.34b) carefully. The equation for \(x\) contains only \(x\)-coordinates, \(x\)-velocities, and \(x\)-accelerations — no \(y\) appears anywhere. Similarly, the equation for \(y\) contains only \(y\)-quantities — no \(x\) shows up. The two equations are completely independent of each other.
This is a powerful and important fact.
The immediate benefit is practical: a difficult-looking 2D problem becomes two easier 1D problems, each of which we already know how to solve from the previous chapter on straight-line motion.
We shall put this principle to use immediately in the next section (Section 3.9), where a projectile experiences a constant downward acceleration due to gravity. The independence of directions lets us split its motion into a horizontal motion (constant velocity) and a vertical motion (uniform downward acceleration), and analyse the two separately.
3.9 Projectile Motion
As a first, concrete application of the ideas developed in Section 3.8, we consider the motion of a projectile.
A cricket ball hit for a six, a football sent into a corner kick, a javelin arcing across an athletic field, a stone flung from a slingshot — all become projectiles once they leave contact with whatever set them in motion.
The key observation that makes projectile motion tractable is this: after the projectile has left the launcher, the only force acting on it is gravity, which points vertically downward and has the same magnitude everywhere near Earth’s surface. That means the projectile is a body moving in a plane with a constant acceleration — exactly the situation of Section 3.8. So we can use all the results we derived there.
Setting up the problem
We shall assume, throughout this discussion, that air resistance has negligible effect on the motion of the projectile. Suppose the projectile is launched with initial velocity \(\mathbf{v_o}\) making an angle \(\theta_o\) with the horizontal x-axis, as shown in Fig. 3.16.
Figure to come
Fig. 3.16 – A projectile launched from origin O at an angle θ_o above the horizontal x-axis. The initial velocity v_o is shown as an arrow, with its horizontal component (v_o cos θ_o) along x and vertical component (v_o sin θ_o) along y indicated. Gravity is marked as a = −g ĵ pointing downward. The full parabolic trajectory is sketched leading up and back down.
After the object has been projected, the only acceleration acting on it is gravity, pointing vertically downward:
\[\mathbf{a} = -g\,\hat{\mathbf{j}}\]
so that
\[a_x = 0, \qquad a_y = -g \quad (3.35)\]
The components of the initial velocity \(\mathbf{v_o}\) are
\[v_{ox} = v_o \cos\theta_o\] \[v_{oy} = v_o \sin\theta_o \quad (3.36)\]
If we take the initial position (the launch point) to be the origin of our reference frame, then \(x_o = 0\) and \(y_o = 0\). With these substitutions, Eq. (3.34b) — the position equation for constant acceleration — becomes
\[x = (v_o \cos\theta_o)\, t\] \[y = (v_o \sin\theta_o)\, t - \tfrac{1}{2}\,g\, t^2 \quad (3.37)\]
And the velocity components at time \(t\), from Eq. (3.33b), are
\[v_x = v_{ox} = v_o \cos\theta_o\] \[v_y = v_o \sin\theta_o - g\, t \quad (3.38)\]
Horizontal and vertical motions are independent
Look carefully at Eqs. (3.37) and (3.38). The x-motion involves only a constant velocity \(v_o \cos\theta_o\) — no acceleration, no changing speed in that direction. The y-motion, on the other hand, is exactly that of an object thrown vertically upward at speed \(v_o \sin\theta_o\) and pulled back down by gravity. The two coexist without interfering with each other.
The choice of mutually perpendicular axes has thus cleanly separated projectile motion into two independent problems: - Horizontal: constant velocity, no acceleration. - Vertical: initial upward velocity, constant downward acceleration \(g\) — just like free fall.
This is shown graphically in Fig. 3.17. At every point along the arc, the horizontal component of velocity has the same value, while the vertical component changes — decreasing on the way up, becoming zero at the topmost point, then reversing sign and growing on the way down.
Figure to come
Fig. 3.17 – A parabolic trajectory drawn from origin O. Velocity vectors v shown at four points along the path (launch, rising, peak, descending, and landing). At each point, the horizontal component v_x î (always the same length) and vertical component v_y ĵ (varying in length and sign) are drawn as decomposition arrows. At the peak, v_y = 0 is marked, so v at the peak is purely horizontal.
At the point of maximum height, \(v_y = 0\), so the angle that the instantaneous velocity makes with the horizontal is \(\theta = \tan^{-1}(v_y/v_x) = 0\) — the velocity there is purely horizontal.
Equation of the path — a parabola
What shape does the projectile’s path actually trace out? To find out, we eliminate the time \(t\) between the two expressions for \(x\) and \(y\) in Eq. (3.37).
From the first equation, \(t = x / (v_o \cos\theta_o)\). Substituting into the second,
\[y = (\tan\theta_o)\, x - \frac{g}{2(v_o \cos\theta_o)^2}\, x^2 \quad (3.39)\]
Since \(g\), \(\theta_o\), and \(v_o\) are all constants, Eq. (3.39) has the form \(y = ax + bx^2\) — the equation of a parabola. So the trajectory of a projectile, in the absence of air resistance, is exactly a parabola (Fig. 3.17).
Time of maximum height
How long does the projectile take to reach the highest point of its arc? Call this time \(t_m\). At the peak, the vertical velocity momentarily vanishes: \(v_y = 0\). Setting Eq. (3.38) to zero,
\[v_o \sin\theta_o - g\, t_m = 0\]
\[t_m = \frac{v_o \sin\theta_o}{g} \quad (3.40a)\]
Time of flight
The total time the projectile spends in the air is called the time of flight, \(T_f\). It is the total time from launch (at \(y = 0\)) until the projectile returns to the same level (\(y = 0\)).
Setting \(y = 0\) in Eq. (3.37) and discarding the trivial \(t = 0\) solution,
\[T_f = \frac{2\, v_o \sin\theta_o}{g} \quad (3.40b)\]
Notice that \(T_f = 2\, t_m\), which is exactly what the symmetry of the parabola tells us — the ascent and the descent take equal times.
Maximum height
The maximum height \(h_m\) reached by the projectile is the value of \(y\) at \(t = t_m\). Substituting into Eq. (3.37),
\[h_m = (v_o \sin\theta_o)\!\left(\frac{v_o \sin\theta_o}{g}\right) - \frac{g}{2}\!\left(\frac{v_o \sin\theta_o}{g}\right)^{\!2}\]
\[h_m = \frac{(v_o \sin\theta_o)^2}{2g} \quad (3.41)\]
Only the vertical component of the launch velocity matters for how high the projectile goes.
Horizontal range
The horizontal range, \(R\), is the horizontal distance travelled from the launch point (\(x = y = 0\)) to the landing point where the projectile returns to \(y = 0\). Since the horizontal motion has constant velocity \(v_o \cos\theta_o\), the range is that velocity multiplied by the time of flight:
\[R = (v_o \cos\theta_o)\, T_f = (v_o \cos\theta_o) \cdot \frac{2\, v_o \sin\theta_o}{g}\]
Using the identity \(2\sin\theta \cos\theta = \sin 2\theta\),
\[R = \frac{v_o^2\, \sin 2\theta_o}{g} \quad (3.42a)\]
For a given launch speed \(v_o\), the range depends only on \(\sin 2\theta_o\), which is maximum when \(2\theta_o = 90°\) — that is, when
\[\theta_o = 45°\]
At that angle, the range takes its largest possible value:
\[R_m = \frac{v_o^2}{g} \quad (3.42b)\]
So a projectile launched with a given speed will fly the farthest when its launch angle is \(45°\) above the horizontal. This is why competitive shot-putters, discus throwers, and long jumpers train to keep their release angles close to (but slightly less than) \(45°\) — the small adjustment comes from air resistance and from the fact that the release height and the landing height are not exactly the same.
3.10 Uniform Circular Motion
When an object moves along a circular path at a constant speed, the motion is called uniform circular motion. The word “uniform” refers to the speed, which stays the same throughout — not to the velocity, which as we shall see is constantly changing.
Consider an object moving at uniform speed \(v\) in a circle of radius \(R\), as shown in Fig. 3.18. Even though its speed is unchanging, its direction of motion is turning at every instant. And any change in direction is a change in velocity — so the object is accelerating, even though it is not speeding up or slowing down.
This is a striking idea worth pausing over. In one-dimensional motion, “no change in speed” and “no acceleration” mean the same thing. In two-dimensional motion, they do not. A body can move at a perfectly constant speed and still have a non-zero acceleration, provided its direction is changing.
Our job in this section is to find the magnitude and direction of this acceleration.
Direction of the acceleration
Let \(\mathbf{r}\) and \(\mathbf{r'}\) be the position vectors of the object when it is at points P and P′ on the circle, and let \(\mathbf{v}\) and \(\mathbf{v'}\) be the corresponding velocities. Since velocity is always tangent to the path (Section 3.7), \(\mathbf{v}\) is perpendicular to \(\mathbf{r}\), and \(\mathbf{v'}\) is perpendicular to \(\mathbf{r'}\).
Figure to come
Fig. 3.18(a) – A circle centred at C; two positions P and P’ on the circle joined to C by position vectors r and r’ with angle Δθ at C; the chord Δr from P to P’ drawn. Adjacent small panels: (a1) shows tangent velocity vectors v and v’ at P and P’ pointing along the direction of motion; (a2) shows a triangle GHI where v and v’ are drawn from a common point with Δv = v’ − v as the closing side, angled inward toward the centre.
Draw the two velocity vectors from a common point [Fig. 3.18(a2)] and complete a triangle with the vector \(\Delta \mathbf{v} = \mathbf{v'} - \mathbf{v}\) as the closing side. Because \(\mathbf{v} \perp \mathbf{r}\) and \(\mathbf{v'} \perp \mathbf{r'}\), the change \(\Delta \mathbf{v}\) turns out to be perpendicular to the chord \(\Delta \mathbf{r}\) joining P to P′.
Since the average acceleration is \(\bar{\mathbf{a}} = \Delta \mathbf{v} / \Delta t\), and \(\Delta t\) is a positive scalar, \(\bar{\mathbf{a}}\) points in the same direction as \(\Delta \mathbf{v}\) — which is perpendicular to \(\Delta \mathbf{r}\). If we place \(\Delta \mathbf{v}\) on the line bisecting the angle between \(\mathbf{r}\) and \(\mathbf{r'}\), we see it points inward — towards the centre C of the circle.
Figure to come
Fig. 3.18(b) – Same construction as Fig. 3.18(a) but with a smaller angle Δθ (P and P’ closer together); again Δv is perpendicular to the chord Δr and points inward toward the centre.
Now shrink \(\Delta t\) towards zero. The point P′ slides towards P, and the average acceleration becomes the instantaneous acceleration at P. At every stage of this shrinking process, \(\Delta \mathbf{v}\) has been pointing inward. In the limit, it points exactly along the radius, from P towards the centre C [Fig. 3.18(c)].
Figure to come
Fig. 3.18(c) – The circle with a single point P on it; the instantaneous acceleration a drawn at P as an arrow pointing from P to the centre C along the radius; velocity v at P shown as a tangent to the circle.
This acceleration is called centripetal acceleration, from Latin roots meaning “centre-seeking.” The term was introduced by Newton in the seventeenth century.
Magnitude of the centripetal acceleration
By definition,
\[|\mathbf{a}| = \lim_{\Delta t \to 0}\frac{|\Delta \mathbf{v}|}{\Delta t}\]
To evaluate this limit, we use an elegant geometric observation. Let \(\Delta\theta\) be the angle between the position vectors \(\mathbf{r}\) and \(\mathbf{r'}\) at C. Since the velocity vectors \(\mathbf{v}\) and \(\mathbf{v'}\) are always perpendicular to the corresponding position vectors, the angle between \(\mathbf{v}\) and \(\mathbf{v'}\) is also \(\Delta\theta\). So the isosceles triangle CPP′ formed by \(\mathbf{r}\), \(\mathbf{r'}\), and \(\Delta \mathbf{r}\) is similar to the isosceles triangle GHI formed by \(\mathbf{v}\), \(\mathbf{v'}\), and \(\Delta \mathbf{v}\).
Similar triangles have equal ratios of base to side:
\[\frac{|\Delta \mathbf{v}|}{v} = \frac{|\Delta \mathbf{r}|}{R}\]
So
\[|\Delta \mathbf{v}| = \frac{v\, |\Delta \mathbf{r}|}{R}\]
Substituting into the acceleration limit,
\[|\mathbf{a}| = \lim_{\Delta t \to 0}\frac{|\Delta \mathbf{v}|}{\Delta t} = \lim_{\Delta t \to 0}\frac{v\, |\Delta \mathbf{r}|}{R\, \Delta t} = \frac{v}{R}\lim_{\Delta t \to 0}\frac{|\Delta \mathbf{r}|}{\Delta t}\]
For very small \(\Delta t\), the chord PP′ (of length \(|\Delta \mathbf{r}|\)) is almost equal to the arc PP′ (of length \(v\, \Delta t\)). So
\[|\Delta \mathbf{r}| \cong v\, \Delta t, \qquad \frac{|\Delta \mathbf{r}|}{\Delta t} \cong v\]
Taking the limit,
\[\lim_{\Delta t \to 0}\frac{|\Delta \mathbf{r}|}{\Delta t} = v\]
Substituting back,
\[a_c = \frac{v}{R} \cdot v = \frac{v^2}{R} \quad (3.43)\]
So the centripetal acceleration of an object moving at speed \(v\) in a circle of radius \(R\) has magnitude \(v^2/R\), and is always directed towards the centre.
Because \(v\) and \(R\) are both constant, the magnitude of the centripetal acceleration is constant. But its direction rotates as the object moves round the circle. So although its magnitude is fixed, centripetal acceleration is not a constant vector — it turns as the object turns.
Angular description
There is another, often more convenient, way to describe uniform circular motion — using angles instead of distances. As the object moves from P to P′ in time \(\Delta t\), the radius CP sweeps through an angle \(\Delta\theta\), called the angular distance (or angular displacement).
\[\omega = \frac{\Delta\theta}{\Delta t} \quad (3.44)\]
Its SI unit is the radian per second (rad s\(^{-1}\)).
The distance travelled along the arc during the same time interval is \(\Delta s = R\,\Delta\theta\) — the basic relation between arc length, radius, and angle (in radians). So the linear speed is
\[v = \frac{\Delta s}{\Delta t} = R\,\frac{\Delta\theta}{\Delta t} = R\,\omega\]
\[v = R\,\omega \quad (3.45)\]
Substituting into Eq. (3.43), the centripetal acceleration can also be written purely in terms of \(\omega\):
\[a_c = \frac{v^2}{R} = \frac{(R\omega)^2}{R} = \omega^2 R\]
\[a_c = \omega^2 R \quad (3.46)\]
Time period and frequency
The time period \(T\) is the time taken for one complete revolution. The frequency \(\nu = 1/T\) is the number of revolutions per second, measured in hertz (Hz).
In one revolution, the distance travelled is the circumference \(2\pi R\). So
\[v = \frac{2\pi R}{T} = 2\pi R\,\nu \quad (3.47)\]
And in terms of frequency,
\[\omega = 2\pi\nu, \qquad v = 2\pi R\nu, \qquad a_c = 4\pi^2 \nu^2 R \quad (3.48)\]
These are the same physical facts as before, rewritten in the form most useful when the motion is specified by how often the object goes round (say, in “revolutions per minute”).
3.11 Summary
Scalar quantities are quantities with magnitude only. Examples: distance, speed, mass, temperature. They obey the rules of ordinary algebra.
Vector quantities are quantities with both magnitude and direction. Examples: displacement, velocity, acceleration, force. They obey the special rules of vector algebra — the triangle law or, equivalently, the parallelogram law of addition.
A vector \(\mathbf{A}\) multiplied by a real number \(\lambda\) gives a vector \(\lambda\mathbf{A}\) of magnitude \(|\lambda|\,|\mathbf{A}|\), whose direction is the same as \(\mathbf{A}\) if \(\lambda > 0\) and opposite to \(\mathbf{A}\) if \(\lambda < 0\).
Two vectors \(\mathbf{A}\) and \(\mathbf{B}\) may be added graphically using the head-to-tail (triangle) method or the parallelogram method. Both give the same resultant.
Vector addition is commutative:
\[\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}\]
and associative:
\[(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})\]
- A null vector (or zero vector) \(\mathbf{0}\) has zero magnitude. Since its magnitude is zero, its direction is not specified. Its properties are:
\[\mathbf{A} + \mathbf{0} = \mathbf{A}, \qquad \lambda\,\mathbf{0} = \mathbf{0}, \qquad 0\,\mathbf{A} = \mathbf{0}\]
- Subtraction of one vector from another is defined as the addition of its negative:
\[\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})\]
- A vector \(\mathbf{A}\) can be resolved into two component vectors along any two chosen non-collinear vectors \(\mathbf{a}\) and \(\mathbf{b}\) in the same plane:
\[\mathbf{A} = \lambda\mathbf{a} + \mu\mathbf{b}\]
where \(\lambda\) and \(\mu\) are real numbers.
- A unit vector has magnitude 1 and specifies a direction only. Along any vector \(\mathbf{A}\), the unit vector is
\[\hat{\mathbf{n}} = \frac{\mathbf{A}}{|\mathbf{A}|}\]
The unit vectors \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), \(\hat{\mathbf{k}}\) point along the x-, y-, and z-axes of a right-handed rectangular coordinate system.
- In two dimensions, any vector \(\mathbf{A}\) can be written as
\[\mathbf{A} = A_x\hat{\mathbf{i}} + A_y\hat{\mathbf{j}}\]
where \(A_x\) and \(A_y\) are its components along the x- and y-axes. If \(\mathbf{A}\) makes an angle \(\theta\) with the x-axis, then \(A_x = A\cos\theta\), \(A_y = A\sin\theta\), and
\[A = |\mathbf{A}| = \sqrt{A_x^2 + A_y^2}, \qquad \tan\theta = \frac{A_y}{A_x}\]
- Two vectors can be conveniently added by the analytical method. If \(\mathbf{R} = \mathbf{A} + \mathbf{B}\) in the x-y plane, then
\[\mathbf{R} = R_x\hat{\mathbf{i}} + R_y\hat{\mathbf{j}}\]
where \(R_x = A_x + B_x\) and \(R_y = A_y + B_y\). In three dimensions, a third component along \(\hat{\mathbf{k}}\) is added in the same way.
- The position vector of an object in the x-y plane is \(\mathbf{r} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}}\), and the displacement from position \(\mathbf{r}\) to position \(\mathbf{r'}\) is
\[\Delta\mathbf{r} = \mathbf{r'} - \mathbf{r} = (x' - x)\hat{\mathbf{i}} + (y' - y)\hat{\mathbf{j}} = \Delta x\,\hat{\mathbf{i}} + \Delta y\,\hat{\mathbf{j}}\]
- If an object undergoes a displacement \(\Delta\mathbf{r}\) in time \(\Delta t\), its average velocity is \(\overline{\mathbf{v}} = \Delta\mathbf{r}/\Delta t\). The (instantaneous) velocity at time \(t\) is the limiting value of the average velocity as \(\Delta t \to 0\):
\[\mathbf{v} = \lim_{\Delta t \to 0}\frac{\Delta\mathbf{r}}{\Delta t} = \frac{d\mathbf{r}}{dt}\]
In component form,
\[\mathbf{v} = v_x\hat{\mathbf{i}} + v_y\hat{\mathbf{j}} + v_z\hat{\mathbf{k}}, \qquad v_x = \frac{dx}{dt},\quad v_y = \frac{dy}{dt},\quad v_z = \frac{dz}{dt}\]
The velocity at any instant is always tangent to the path at that point.
- If the velocity of an object changes from \(\mathbf{v}\) to \(\mathbf{v'}\) in time \(\Delta t\), its average acceleration is \(\overline{\mathbf{a}} = \Delta\mathbf{v}/\Delta t\). The (instantaneous) acceleration is the limit as \(\Delta t \to 0\):
\[\mathbf{a} = \lim_{\Delta t \to 0}\frac{\Delta\mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}\]
In component form,
\[\mathbf{a} = a_x\hat{\mathbf{i}} + a_y\hat{\mathbf{j}} + a_z\hat{\mathbf{k}}, \qquad a_x = \frac{dv_x}{dt},\quad a_y = \frac{dv_y}{dt},\quad a_z = \frac{dv_z}{dt}\]
- For an object moving in a plane with constant acceleration \(\mathbf{a}\) (magnitude \(a = \sqrt{a_x^2 + a_y^2}\)), starting at position \(\mathbf{r_o}\) with initial velocity \(\mathbf{v_o}\) at \(t = 0\), the position and velocity at any later time \(t\) are:
\[\mathbf{r} = \mathbf{r_o} + \mathbf{v_o}\,t + \tfrac{1}{2}\,\mathbf{a}\,t^2\]
\[\mathbf{v} = \mathbf{v_o} + \mathbf{a}\,t\]
In component form,
\[x = x_o + v_{ox}t + \tfrac{1}{2}a_x t^2, \qquad v_x = v_{ox} + a_x t\]
\[y = y_o + v_{oy}t + \tfrac{1}{2}a_y t^2, \qquad v_y = v_{oy} + a_y t\]
Motion in a plane can therefore be treated as the superposition of two simultaneous one-dimensional motions along two perpendicular directions.
- A projectile is any object in flight under the sole action of gravity. If launched from the origin with initial velocity \(\mathbf{v_o}\) at an angle \(\theta_o\) above the horizontal, then at time \(t\):
\[x = (v_o\cos\theta_o)\,t\]
\[y = (v_o\sin\theta_o)\,t - \tfrac{1}{2}g\,t^2\]
\[v_x = v_o\cos\theta_o, \qquad v_y = v_o\sin\theta_o - g\,t\]
The path is a parabola, given by
\[y = (\tan\theta_o)\,x - \frac{g\,x^2}{2(v_o\cos\theta_o)^2}\]
The maximum height, time to reach that height, and horizontal range are:
\[h_m = \frac{(v_o\sin\theta_o)^2}{2g}, \qquad t_m = \frac{v_o\sin\theta_o}{g}, \qquad R = \frac{v_o^2\,\sin 2\theta_o}{g}\]
The range \(R\) is maximum at a launch angle of \(\theta_o = 45°\).
- When an object moves along a circular path at constant speed, its motion is called uniform circular motion. The magnitude of its (centripetal) acceleration is
\[a_c = \frac{v^2}{R}\]
and its direction is always towards the centre of the circle. The angular speed \(\omega\) is related to the linear speed by \(v = \omega R\), and the acceleration can also be written as
\[a_c = \omega^2 R\]
If \(T\) is the time period of revolution and \(\nu\) is the frequency, then \(\omega = 2\pi\nu\), \(v = 2\pi R\nu\), and
\[a_c = 4\pi^2 \nu^2 R\]
Table of key physical quantities
| Physical Quantity | Symbol | Dimensions | Unit | Remark |
|---|---|---|---|---|
| Position vector | \(\mathbf{r}\) | \([\text{L}]\) | m | Vector; may be denoted by any other symbol |
| Displacement | \(\Delta\mathbf{r}\) | \([\text{L}]\) | m | — do — |
| Average velocity | \(\overline{\mathbf{v}}\) | \([\text{LT}^{-1}]\) | m s\(^{-1}\) | \(= \Delta\mathbf{r}/\Delta t\), vector |
| Instantaneous velocity | \(\mathbf{v}\) | \([\text{LT}^{-1}]\) | m s\(^{-1}\) | \(= d\mathbf{r}/dt\), vector |
| Average acceleration | \(\overline{\mathbf{a}}\) | \([\text{LT}^{-2}]\) | m s\(^{-2}\) | \(= \Delta\mathbf{v}/\Delta t\), vector |
| Instantaneous acceleration | \(\mathbf{a}\) | \([\text{LT}^{-2}]\) | m s\(^{-2}\) | \(= d\mathbf{v}/dt\), vector |
| Time of maximum height (projectile) | \(t_m\) | \([\text{T}]\) | s | \(= v_o\sin\theta_o / g\) |
| Maximum height (projectile) | \(h_m\) | \([\text{L}]\) | m | \(= (v_o\sin\theta_o)^2 / 2g\) |
| Horizontal range (projectile) | \(R\) | \([\text{L}]\) | m | \(= v_o^2\,\sin 2\theta_o / g\) |
| Angular speed (circular motion) | \(\omega\) | \([\text{T}^{-1}]\) | rad s\(^{-1}\) | \(= \Delta\theta/\Delta t = v/R\) |
| Centripetal acceleration (circular motion) | \(a_c\) | \([\text{LT}^{-2}]\) | m s\(^{-2}\) | \(= v^2/R = \omega^2 R\) |
3.12 Points to Ponder
Path length vs displacement. The path length that an object actually traverses between two points is, in general, not the same as the magnitude of its displacement. Displacement depends only on the two endpoints — where the object started and where it ended. Path length depends on the entire route taken. The two are equal only when the object moves in a straight line without ever reversing its direction. In every other case (including all curved motion), the path length is greater than the magnitude of the displacement.
Average speed vs magnitude of average velocity. Following from Point 1, the average speed of an object is always greater than or equal to the magnitude of its average velocity over the same time interval. The two are equal only when the path length equals the magnitude of the displacement — that is, only for straight-line motion with no direction reversal. A car that drives around a large loop and returns to its starting point has a positive average speed but zero average velocity.
The vector equations \(\mathbf{v} = \mathbf{v_0} + \mathbf{a}\,t\) and \(\mathbf{r} = \mathbf{r_0} + \mathbf{v_0}\,t + \tfrac{1}{2}\mathbf{a}\,t^2\) do not involve any choice of axes. Equations (3.33a) and (3.34a) hold as vector equations, independent of any coordinate system. This is a real advantage: you can pick whatever pair (or triple) of independent axes is most convenient for your problem and resolve both sides of the vector equation along those axes. In projectile problems the horizontal-and-vertical choice is natural; in other problems, a rotated pair of axes may be far easier.
The uniform-acceleration equations do not apply to uniform circular motion. In uniform circular motion, the magnitude of the acceleration is constant, but its direction changes at every instant — the acceleration vector rotates. Because Eqs. (3.33a) and (3.34a) assume \(\mathbf{a}\) is a constant vector (same magnitude and same direction throughout), they cannot be applied to uniform circular motion. Trying to use them here is one of the most common exam mistakes.
Resultant velocity vs relative velocity — do not confuse them. If an object is subjected simultaneously to two velocities \(\mathbf{v_1}\) and \(\mathbf{v_2}\) (for example, a swimmer with velocity \(\mathbf{v_1}\) in a river with current \(\mathbf{v_2}\)), its resultant velocity as seen from the ground is \(\mathbf{v} = \mathbf{v_1} + \mathbf{v_2}\). This is a sum. But the velocity of object 1 as seen from object 2 — the velocity of 1 relative to 2 — is \(\mathbf{v_{12}} = \mathbf{v_1} - \mathbf{v_2}\). This is a difference. In both formulas, \(\mathbf{v_1}\) and \(\mathbf{v_2}\) must be measured in the same common reference frame. Mixing up resultant with relative is one of the most frequent errors in problem-solving.
Centripetal acceleration points to the centre only when the speed is constant. The result that the acceleration in circular motion is directed towards the centre applies specifically to uniform circular motion. If the speed itself is changing as the object goes round the circle (non-uniform circular motion), the acceleration also has a tangential component along the direction of motion, and the total acceleration no longer points purely towards the centre.
The trajectory of an object depends on both the acceleration and the initial conditions. Acceleration alone does not determine the path an object will trace out. The initial position and the initial velocity are equally important. Under the very same acceleration due to gravity, an object dropped from rest falls in a straight vertical line, an object thrown horizontally traces a parabola, and an object launched vertically upward goes straight up and straight back down. Same acceleration, three different trajectories — the difference is entirely in the initial conditions.
3.13 NCERT Questions
State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Pick out the only vector quantity in the following list: temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful: (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.
Read each statement below carefully and state with reasons, if it is true or false: (a) The magnitude of a vector is always a scalar. (b) Each component of a vector is always a scalar. (c) The total path length is always equal to the magnitude of the displacement vector of a particle. (d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater than or equal to the magnitude of the average velocity of the particle over the same interval of time. (e) Three vectors not lying in a plane can never add up to give a null vector.
Establish the following vector inequalities geometrically or otherwise:
- \(|\mathbf{a}+\mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}|\)
- \(|\mathbf{a}+\mathbf{b}| \geq ||\mathbf{a}| - |\mathbf{b}||\)
- \(|\mathbf{a}-\mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}|\)
- \(|\mathbf{a}-\mathbf{b}| \geq ||\mathbf{a}| - |\mathbf{b}||\) When does the equality sign above apply?
Given \(\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = 0\), which of the following statements are correct: (a) \(\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}\) must each be a null vector. (b) The magnitude of \((\mathbf{a} + \mathbf{c})\) equals the magnitude of \((\mathbf{b} + \mathbf{d})\). (c) The magnitude of \(\mathbf{a}\) can never be greater than the sum of the magnitudes of \(\mathbf{b}, \mathbf{c}, \mathbf{d}\). (d) \(\mathbf{b} + \mathbf{c}\) must lie in the plane of \(\mathbf{a}\) and \(\mathbf{d}\) if \(\mathbf{a}\) and \(\mathbf{d}\) are not collinear, and in the line of \(\mathbf{a}\) and \(\mathbf{d}\) if they are collinear.
Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skated?
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?
On an open ground, a motorist follows a track that turns to his left by an angle of \(60°\) after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth, and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of the average velocity? Are the two equal?
The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of \(40\ \text{m s}^{-1}\) can cover without hitting the ceiling of the hall?
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How high above the ground can the cricketer throw the same ball?
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of the acceleration of the stone?
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Read each statement below carefully and state, with reasons, if it is true or false: (a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre. (b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point. (c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
The position of a particle is given by \[\mathbf{r} = 3.0\,t\,\hat{\mathbf{i}} - 2.0\,t^2\,\hat{\mathbf{j}} + 4.0\,\hat{\mathbf{k}}\ \text{m}\] where \(t\) is in seconds and the coefficients have the proper units for \(\mathbf{r}\) to be in metres. (a) Find the \(\mathbf{v}\) and \(\mathbf{a}\) of the particle. (b) What is the magnitude and direction of the velocity of the particle at \(t = 2.0\) s?
A particle starts from the origin at \(t = 0\) s with a velocity of \(10.0\,\hat{\mathbf{j}}\) m/s and moves in the x-y plane with a constant acceleration of \((8.0\,\hat{\mathbf{i}} + 2.0\,\hat{\mathbf{j}})\ \text{m s}^{-2}\). (a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time? (b) What is the speed of the particle at that time?
\(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) are unit vectors along the x- and y-axis respectively. What is the magnitude and direction of the vectors \(\hat{\mathbf{i}} + \hat{\mathbf{j}}\), and \(\hat{\mathbf{i}} - \hat{\mathbf{j}}\)? What are the components of a vector \(\mathbf{A} = 2\,\hat{\mathbf{i}} + 3\,\hat{\mathbf{j}}\) along the directions of \(\hat{\mathbf{i}} + \hat{\mathbf{j}}\) and \(\hat{\mathbf{i}} - \hat{\mathbf{j}}\)? [You may use the graphical method.]
For any arbitrary motion in space, which of the following relations are true: (a) \(\mathbf{v}_\text{average} = (1/2)\,[\mathbf{v}(t_1) + \mathbf{v}(t_2)]\); (b) \(\mathbf{v}_\text{average} = [\mathbf{r}(t_2) - \mathbf{r}(t_1)] / (t_2 - t_1)\); (c) \(\mathbf{v}(t) = \mathbf{v}(0) + \mathbf{a}\,t\); (d) \(\mathbf{r}(t) = \mathbf{r}(0) + \mathbf{v}(0)\,t + (1/2)\mathbf{a}\,t^2\); (e) \(\mathbf{a}_\text{average} = [\mathbf{v}(t_2) - \mathbf{v}(t_1)] / (t_2 - t_1)\). (The ‘average’ stands for the average of the quantity over the time interval \(t_1\) to \(t_2\).)
Read each statement below carefully and state, with reasons and examples, if it is true or false. A scalar quantity is one that (a) is conserved in a process, (b) can never take negative values, (c) must be dimensionless, (d) does not vary from one point to another in space, (e) has the same value for observers with different orientations of axes.
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is \(30°\), what is the speed of the aircraft?
3.14 Check Your Concepts
In one-dimensional motion, the velocity and acceleration of an object are always either parallel or anti-parallel to each other. Explain why this restriction does not apply to motion in two or three dimensions, and give one everyday example of motion where the velocity and acceleration are perpendicular.
A student writes: “Since the horizontal velocity of a projectile does not change during its flight, no horizontal force acts on it.” Is this reasoning fully correct? In your answer, state what force actually acts on the projectile during its flight and explain why the horizontal component of velocity nevertheless stays constant in the idealized case studied in this chapter.
An object moves along a circular track with steadily increasing speed. Is its acceleration still directed purely towards the centre of the track? If not, describe the two components of the acceleration and the physical meaning of each.
A student computes the components of a vector making an angle of \(240°\) with the positive x-axis (measured anticlockwise) and finds both \(A_x\) and \(A_y\) to be negative. Explain, without doing any calculation, why this outcome is physically expected.
A ball is thrown from the ground at \(30°\) above the horizontal and travels a certain horizontal range \(R\) on Earth. If the same ball were thrown at the same angle and speed on the surface of the Moon (where the acceleration due to gravity is roughly \(g/6\)), would the range be smaller, the same, or larger? What about at \(60°\)? Explain briefly.
Centripetal acceleration is often described in words as “acceleration directed towards the centre.” Why, then, does an object in uniform circular motion not actually spiral inward and reach the centre? Explain in terms of the direction of the velocity at each instant.
Two students are asked whether a null vector has a direction. Student A says: “Yes, because every vector has a direction.” Student B says: “No, because a vector of zero magnitude has no direction that can be defined.” Which student is correct according to the definition used in this chapter, and why?
3.15 Practice with Numericals
Vector \(\mathbf{A}\) has magnitude 5 units at \(37°\) above the positive x-axis, and vector \(\mathbf{B}\) has magnitude 8 units at \(143°\) above the positive x-axis (both measured anticlockwise). Using the analytical method, find the magnitude and direction of \(\mathbf{A} + \mathbf{B}\).
A ball is thrown from ground level at an angle of \(60°\) above the horizontal with a speed of \(20\ \text{m s}^{-1}\). Taking \(g = 10\ \text{m s}^{-2}\), find (a) the maximum height, (b) the time of flight, (c) the horizontal range, and (d) the velocity (magnitude and direction) of the ball at the moment it lands back at the same height.
A boat can travel at \(5\ \text{km h}^{-1}\) in still water. A river flows east to west at \(3\ \text{km h}^{-1}\). The boat is pointed directly across the river (perpendicular to the current). Find (a) the boat’s actual velocity as seen from the shore (magnitude and direction), (b) the angle its actual path makes with the intended straight-across direction, and (c) if the river is 400 m wide, how far downstream the boat drifts by the time it reaches the opposite bank.
A particle moves in the x-y plane with its coordinates given by \(x(t) = 2t + t^2\) and \(y(t) = 3t - t^2\), both in metres and \(t\) in seconds. Find (a) the velocity vector at \(t = 2\) s, (b) the acceleration vector at any time \(t\), and (c) the speed of the particle at \(t = 2\) s.
A stone is tied to a string of length \(1.0\) m and whirled in a horizontal circle. The string can safely withstand a maximum centripetal acceleration of \(40\ \text{m s}^{-2}\). What is the maximum linear speed the stone can have before the string breaks? What is the corresponding angular speed in revolutions per minute?
A missile is launched from level ground at an angle of \(30°\) above the horizontal with an initial speed of \(500\ \text{m s}^{-1}\). Taking \(g = 9.8\ \text{m s}^{-2}\) and neglecting air resistance, calculate (a) the time of flight, (b) the horizontal range, and (c) the maximum height reached.
A car moving at \(20\ \text{m s}^{-1}\) enters a horizontal circular curve of radius 100 m. (a) What is the magnitude of its centripetal acceleration? (b) If the driver doubles the speed to \(40\ \text{m s}^{-1}\) on the same curve, by what factor does the centripetal acceleration change?
A helicopter is hovering at rest at a height of 200 m above the ground. From this hovering helicopter, one package is dropped from rest and, at the same instant, a second package is thrown horizontally with a speed of \(30\ \text{m s}^{-1}\). Taking \(g = 10\ \text{m s}^{-2}\) and neglecting air resistance, find (a) which package hits the ground first, and by how much time (if any), (b) the horizontal distance from the hover point at which the second package lands, and (c) the velocity (magnitude and direction) with which the second package hits the ground.