Physics NEET: Forces and Laws of Motion

Course Content

Total learning: 3 lessons / 4 quizzes

Forces and Motion


A push or pull on a body is called force. Forces are used in our everyday actions like pushing, pulling, lifting, stretching, twisting and pressing. In subsequent heads the effects of force and types of force are described.

Motion is a change in position of object over time. Motion is described in terms of:

  • displacement
  • distance
  • velocity
  • acceleration
  • time and
  • speed.

In following discussions, we learn how to describe motion. The concepts of position, displacement, velocity and acceleration that are needed to describe the motion of an object along a straight line, a plane (two dimensions) or three dimensions (space) are articulated further.


Motion is common to everything in the universe. We walk, run or ride a van. Even when we are sleeping, air move into and out of our lungs and blood flows in arteries and veins. We see leaves falling from trees and water flowing down a dam. Automobiles and planes carry people from one place to the other. The earth rotates once every twenty four hours and revolves round the sun once in a year. The sun itself is in motion in the Milky Way, which is again moving within its local group of galaxies.

Let us see what is motion? As already mentioned in introduction, motion is change in position with time. To further know about motion, it is necessary to understand the concepts of velocity and acceleration. Let us now limit ourselves to the study of motion of objects along a straight line which is also known as rectilinear motion. In the discussions, let us treat the objects in motion as point objects. This approximation is valid so far as the size of the object is much smaller than the distance it moves in a reasonable duration of time. In a good number of situations, in real life, the size of objects can be neglected and they can be considered as point like objects without much error. What causes motion described in this lesson.

Position, Path, Length and Displacement

Earlier we learnt motion is change in position of an object with time. In order to specify position, we need to use a reference point and a set of axes. It is convenient to choose a rectangular coordinate system consisting of three mutually perpendicular axes, labelled X, Y and Z axes. The point of intersection of these three axes is called origin (O) and serves as the reference point. The coordinates (x, y, z) of an object describe the position of the object with respect to coordinate system. To measure time, position a clock in this system. This coordinate system along with a clock constitutes a frame of reference.

If one or more coordinates of an object change with time, we say that the object is in motion. Otherwise, the object is said to be at rest with respect to this frame of reference.

The choice of a set of axes in a frame of reference depends upon the situation. As an example, for describing motion in one direction, we need only one axis. To describe motion in two/three dimensions, we need a set of two/three axes.

Description of an event depends on the frame of reference chosen for the description. For example, when we say that a van is moving on a road, we describe the van with respect to a frame of reference attached to you or to the ground. But with respect to a frame of reference attached with a person sitting on the van, the van is at rest.

To describe motion along a straight line, we can choose an axis, say X-axis, so that it coincides with the path of the object. We then measure the position of the object with reference to a conveniently chosen origin, say O as shown in Fig. 1 below. Positions to the right of O are taken as positive and to the left of O as negative. Following this convention, the position coordinates of point P and Q in Fig .1 are +360 m and +240 m. Similarly, the position coordinate of point R is -120 m.

Fig 1 – X-axis origin and positions of car at different times

Path length

Consider the motion of a car along a straight line. We choose the x-axis such that it coincides with the path of the van’s motion and origin of the axis as the point from where the van started moving. i.e, the van was at x=0 at t=0 (Fig. 1). Let P, Q and R represent the positions of the van at different instants of time. Consider two cases of motion. In the first case, the van moves from O to P. Then the distance moved by the van is OP=+360 m. This distance is called the path length traversed by the van. In the second case, the car moves from O to P and then moves back from P to Q. During this course of motion, the path length traversed is OP + PQ= +360 m + (+120 m) = +480 m. Path length is a scalar quantity-a quantity that has a magnitude only and no direction.


It is useful to define another quantity displacement as the change in position. Let x1 and x2 be the positions of an object at time t1 and t2 . Then its displacement, denoted by Δx, in time Δt = (t2 – t1) is given by the difference between the final and initial positions;

Δx = x2 – x1

(We use Greek letter (Δ) to denote a change in quantity.)

If x2 > x1, Δx is positive; and if x2 < x1, Δx is negative.

Displacement has both magnitude and direction. Such quantities are represented by vectors. You will read about vectors in the next chapter. Presently, we are dealing with motion along a straight line (also called rectilnear motion) only. In one dimensional motion, there are only two directions (backward and forward, upward and downward) in which an object can move and these two directions can easily specified by + and – signs. For example, displacement of the van in moving from O to P is;

Δx = x2 – x1 = (+360 m) – 0 m =+360 m.

The displacement has a magnitude of 360 m and is directed in the positive x direction as indicated by the + sign. Similarly, the displacement of the van from P to Q is 240 m – 360 m = -120 m. The negative sign indicates that the direction displacement. Thus, it is necessary to use vector rotation for discussing motion of objects in one dimension.

The magnitude of displacement may or may not be equal to the path length traversed by an object. For example, for motion of the van from O to P, the path length is +360 m. In this case, the magnitude of displacement (360 m) is equal to the path length (360 m). But consider the motion of the van from O to P and back to Q. In this case, the path length = (+360 m) + (+120 m) = + 480 m. However, the displacement = (+240 m) – (0 m) = + 240 m. Thus, the magnitude of displacement (240 m) is not equal to the path length (480 m).

The magnitude of the displacement for a course of motion may be zero but the corresponding path length is not zero. For example, if the van starts from O, goes to P and then returns to O, the final position coincides with the initial position and the displacement is zero. However, the path length of this journey is OP + PO = 360 m + 360 m = 720 m.

Motion of an object can be represented by a position-time graph as you have already learnt about it. Such a graph is a powerful tool to represent and analyse different aspects of motion of an object. For motion along a straight line, say X-axis, only x-coordinate varies with time and we have an x-t graph. Let us first consider the simple case in which an object is stationary. e.g. a van standing still at x = 40 m. The position-time graph is a straight line parallel to the time axis, as shown in Fig. 2(a).

If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line. Fig. 2(b) shows the position-time graph of such a motion.

Fig 3 – Position time graph of a van

Now, let us consider the motion of a van that starts from rest at time t = 0 s from the origin O and picks up speed till t = 10 s and thereafter moves with uniform speed till t = 18 s. Then the brakes are applied and the van stops at t = 20 s and x = 296 m. The position-time graph for this case is shown in Fig 3. We shall refer to this graph in the following sections.

Average Velocity and Average Speed

When an object is in motion, its position changes with time. But how fast is the position changing with time and in what direction? To describe this, we define the quantity average velocity. Average velocity is defined as the change in position or displacement (Δx) divided by the time intervals (Δt), in which the displacement occurs:

ν = (x2x1)/(t2– t1) = Δx/Δt ……………………………………………………………………….(1)

where x2 and x1 are the positions of the object at time t2 and t1 respectively. Here the bar over the symbol for velocity is a standard notation used in many everyday applications.

Like displacement, average velocity is also a vector quantity. But as explained earlier, for motion in a straight line, the directional aspect of the vector can be taken care of by + and – signs and we do not have to use the vector notation for velocity in this chapter.

Fig 4 – The average velocity is the slope of line P1P2.

Consider the motion of the van in Fig. 3. The portion of the x-t graph between t=0s and t=8s is blown up and shown in Fig. 4. As seen from the plot, the average velocity of the van between time t =5s and t = 7s is:

ν = (x2x1)/( t2– t1) = (27.4-10.0)m/(7-5)s = 8.7 m s-1

Geometrically, this is the slope of the straight line P1P2 connecting the initial position P1 to the final position P2 as shown in Fig. 4.

The average velocity can be positive or negative depending upon the sign of the displacement. It is zero if the displacement is zero. Fig. 5 shows the x-t graphs for an object, moving with positive velocity (Fig. 5a) moving with negative velocity (Fig. 5b) and at rest (Fig. 5c).

Figure 5
Fig. 5

Average velocity as defined involves only the displacement of the object. We have seen earlier that the magnitude of displacement may be different from the actual path length. To describe the rate of motion over the actual path, we introduce another quantity called average speed.

Average speed is defined as the total path traveled divided by the total time interval during which the motion has taken place:

Average speed = Total path length/Total time interval ………………………………..(2)

Average speed has obviously the same unit (m s-1) as that of velocity. But it does not tell us in what direction an object is moving. Thus, it is always positive (in contrast to the average velocity which can be positive or negative). If the motion of an object is along a straight line and in the same direction, the magnitude of displacement is equal to the total path length. In that case, the magnitude of average velocity is equal to the average speed. This is not always the case, as will be seen in the following example.

Example 1
A van is moving along a straight line, say OP in Fig, 1. It moves from O to P in 18 s and returns from P to Q in 6.0 s. What are the average velocity and average speed of the van in going (a) from O to P? and (b) from O to P and back to Q?


(a) Average velocity = Displacement/Time interval

v = +360 m/18 s = +20 m s-1

Average speed = Path length/Time Interval

= 360 m/18 s = 20 m s-1

Thus, in this case the average speed is equal to the magnitude of the average velocity.

(b) In this case,

Average velocity = Displacement/Time interval

v = +240 m/(18 + 6.0) s = +10 m s-1

Average speed = Path length/Time Interval = (OP + PQ)/Δt

= (360+120) m/24 s = 20 m s-1

Thus, in this case the average speed is equal not to the magnitude of the average velocity. This happens because the motion here involves change in direction so that the path length is greater than the magnitude of displacement. This shows that speed is, in general, greater than the magnitude of the velocity.

If the car in Example 1 moves from O to P and comes back to O in the same interval, average speed is 20 m/s but the average velocity is zero.

Instantaneous Velocity and Speed

The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t.

The velocity at an instant is defined as the limit of the average velocity as the time interval Δt becomes infinitesimally small. In other words,


= dx/dt ………………………………………………………………………….3(b)

where the symbol

stands for the operation of taking limit as Δt→0 of the quantity on its right. In the language of calculus, the quantity on the right hand side of Eq. 3(a) is the differential coefficient of x with respect to t and is denoted by dx/dt. It is the rate of change of position with respect to time, at that instant.

We can use Eq. 3(a) for obtaining the value of velocity at an instant either graphically or numerically. Suppose that we want to obtain graphically the velocity at time t = 4 s (point P) for the motion of the car represented in Fig. 3. The figure has been redrawn in Fig. 6 choosing different scales to facilitate the calculation.

Fig 6 – Determining velocity from position-time graph.

Velocity at t = 4s is the slope of the tangent to the graph at that instant.
Let us take Δt = 2 s centered at t = 4 s.
Then, by the definition of the average velocity, the slope of the line P1P2 (Fig. 6) gives the value of Δt from 2 s to 1 s. Then the line P1P2 becomes Q1Q2 and its slope gives the value of the average velocity over the interval 3.5 s to 4.5 s. In the limit Δt →0 the line P1P2 becomes tangent to the position-time curve at the point P and the velocity at t = 4 s is given by the slope of the tangent at that point. It is difficult to show this process graphically. But if we use numerical method to obtain the value of the velocity, the meaning of the limiting process becomes clear. For the graph shown in Fig. 6, x = 0.08 , Table 1 gives the value of Δxt calculated for Δt equal to 2.0 s, 1.0 s, 0.5 s, 0.1 s centered at t = 4.0 s. The second and third columns give the value of t1 = [t – (Δt/2)] and t2 = [t + t/2)] and the fourth and fifth columns give the corresponding values of x. i.e, x(t1) = 0.08 (t1and x(t2) = 0.08 (t2)³. The sixth column lists the difference Δx = x(t2)-x(t1) and the last column gives the ratio of Δx and Δt, i.e., the average velocity corresponding to the value of Δt listed in the first column.

We see from Table-1 that as we decrease the value of Δt from 2.0 s to 0.010 s, the value of the average velocity approaches the limiting value of 3.84 m s-1 which is the value of average velocity at t = 4.0 s, i.e, the value of dx/dt at t=4.0 s. In this manner, we can calculate velocity at each instant for motion of the car shown in Fig. 3.

Table 1 – Limiting value of Δx/Δt at t = 4 s

Δt(s) t1(s) t2(s) xt1(m) xt2(m) Δx(m) Δx ⁄ Δt(m s-1)
2.0 3.0 5.0 2.16 10.0 7.84 3.92
1.0 3.5 4.5 3.43 7.29 3.86 3.86
0.5 3.75 4.25 4.21875 6.14125 1.9225 3.845
0.1 3.95 4.05 4.93039 5.31441 0.38402 3.8402
0.01 3.995 4.005 5.100824 5.139224 0.384 3.8400

For this case, the variation of velocity with time is found to be as shown in Fig. 7 below:

The graphical method for the determination of the instantaneous velocity is always not a convenient method. For this, we must carefully plot the position-time graph and calculate the value of average velocity as Δt becomes smaller and smaller. It i easier to calculate the value of velocity at different instants or exact expression for the position as a function of time. Then, we calculate Δx ⁄ Δt from the data for decreasing the value of Δt and find the limiting value as we have done in Table 1 or use differential calculus for the given expression and calculate dx/dt at different instants as done in the following example.

Example 2
The position of an object moving along x-axis is given by x = a +bt² where a = 8.5 m, b = 2.5 m s-2 and t is measured in seconds. What is its velocity at t = o s and t = 2.0 s. What is the average velocity between t = 2.o s and t = 4.0 s?

In notation of differential calculus, the velocity is

v = dx/dt = d/dt(a +bt²) = 2bt = 5.0 m s-1

At t = 0 s, v = 0 m s-1 and at t = 2.0 s, v = 10 m s-1

Average velocity = [x(4.0) – x(2.0)](4.0 -2.0)

= (a + 16b a – 4b) ⁄ 2.0 = 6.0 × b

= 6.0 × 2.5 = 15 m s-1

From Fig. 7, we note that during the period t = 10 s to 18 s the velocity is constant. Between period t = 18 s to t = 20 s, it is uniformly decreasing and during the period t = 0 s to t = 10 s, it is increasing. Note that for uniform motion, velocity is same as the average velocity at all instants.

Instantaneous speed or simply speed is the magnitude of velocity. For example, a velocity of 24.0 m s-1 and a velocity of -24.0 m s-1 -both have an associated speed of 24.0 m s-1. It should be noted that though average speed over a finite interval of time is greater or equal to the magnitude of the instantaneous velocity at that instant. Why so?


The velocity of an object, in general, changes during its course of motion. How we can describe this change? Should it be described as the rate of change in velocity with distance or with time ? This was a problem even in Galileo’s time. It was first thought that this change can be explained by the rate of velocity with distance. But, through his studies of motion of freely falling objects and motion of objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is a constant of motion for all objects in free fall. On the other hand, the change in velocity with distance is not constant-It decreases with increasing distance of fall. This led to the concept of acceleration as the rate of change of velocity with time.

The average acceleration ā over a time interval is defined as the change of velocity divided by the time interval :

ā = (v2-v1)/(t2-t1) = Δv/Δt ………………………………………………………(4)

where v2 and v1 are the instantaneous velocities or simply velocities at a time t2 and t1. It is the average change of velocity per unit time. The SI unit for acceleration is m s-2.

On a plot of velocity versus time, the average acceleration is the slope of the straight line connecting the points corresponding to (v2 , t2) and (v1 , t1). The average acceleration for velocity-time graph shown in Fig. 7 for different time intervals 0 s – 10s, 10 s – 18s, and 18s – 20s are:

0 s – 10 s ā = (24 – 0) m s-1/(20-18)s = 2.4 m s-2

10 s – 18 s ā = (24 – 24) m s-1/(18-10)s = 0 m s-2

18 s – 20 s ā = (0 – 24) m s-1/(20-18)s = – 12 m s-2

Instantaneous acceleration is defined in the same way as the instantaneous velocity:

a = limΔt→0 (Δv/Δt) = dv/dt ………………………………………………………………………..(5)

The acceleration at an instant is the slope of the tangent to the v-t curve at that instant. For the v-t curve shown in Fig. 7, we can obtain acceleration at every instant of time. The resulting a-t curve is shown in Fig. 8. We see that acceleration is nonuniform over the period 0 s to 10 s. It is zero between 10 s and 18 s and is constant with value -12 m s-2 between 18 s and 20 s. When the acceleration is uniform, obviously, it equals the average acceleration over that period.

Since velocity is a quantity having both magnitude and direction, a change in velocity may involve either or both of these factors. Acceleration, therefore, may result from a change in speed (magnitude, a change in direction or changes in both. Like velocity, acceleration can also be positive, negative or zero. Position-time graphs for motion with positive, negative and zero acceleration are shown in in Fig. 9 (a), (b) and (c) respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration. As an exercise, identify in Fig. 3, the regions of the curve that correspond to these three cases.

Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is v0 at t = 0 and v at time t, we have,

ā = (v-v0)/(t-0) or, v = v0 + at ………………………………………………………………………(6)

Fig 9 – Position-time graph for motion with (a) positive acceleration; (b) negative acceleration, and (c) zero acceleration.

Let us see how velocity – time graph looks like for some simple cases. Fig. 10 shows velocity-time graph for motion with constant acceleration for the following cases:

(a) An object is moving in a positive direction with a positive acceleration, for example the motion of the car in Fig. 3 between t = 0 s and t = 10 s.

(b) An object is moving in a positive direction with a negative acceleration, for example the motion of the car in Fig. 3 between t = 18 s and t = 20 s.

(c) An object is moving in a negative direction with a negative acceleration, for example the motion of a car moving from O in Fig. 1 in negative x-direction with increasing speed.

(d) An object is moving in a positive direction till time t1 and then turns back with the same negative acceleration, for example the motion of a car moving from point O to point Q in Fig. 1 till time t1 with decreasing speed and turning back and moving with the same negative acceleration.

An interesting feature of a velocity-time graph for any moving object is that the area under the curve represents the displacement over a given time interval. A general proof of this statement requires use of calculus. We can, however, see that it is

true for the simple case of an object moving with constant velocity u. Its velocity-time graph is as shown in Fig. 11. The v-t curve is a straight line parallel to the time axis and the area under it between t=0 and t = T is the area of the rectangle of height u and base T. Therefore, area = u × T = uT which is the displacement in this time interval. How come in this case an area is equal to a distance? Think! Note the dimensions of quantities on the two co-ordinate axes, and you will arrive at the answer.

Note that the x-t, v-t, and a-t graphs shown in several figures in this chapter have sharp kinks at some points implying that the functions are not differentiable at these points. In any realistic situation, the functions will be differentiable at all points and the graphs will be smooth.

What this means physically is that acceleration and velocity cannot change values abruptly at an instant. Changes are always continuous.

Kinematic Equations for Uniformly Accelerated Motion

For uniformly accelerated motion, we can derive some simple equations that relate displacement (x) , time taken (t), initial velocity (v0), final velocity (v) and acceleration (a). Equation (6) already obtained gives a relation between final and initial velocities and v0 of an object moving with uniform acceleration a:

v = v0 + at ……………………………………………… (6)

This relation is graphically represented in Fig. 12. The area under this curve is:

Area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD

= 1/2 (v-v0)t +v0t

As explained in the previous section, the area under v-t curve represents the displacement. Therefore, the displacement x of the object is:

x = 1/2(v-v0)t + v0t ………………………………….(7)

But v-v0 = a t

Therefore, x = 1/2 a t² + v0t

or, x = v0t + 1/2 a t² ………………………………………………(8)

Equation (7) can also be written as,


Equations (3.9a) and (3.9b) mean that the object has undergone displacement x with an average velocity equal to the arithmetic average of the initial and final velocities.

From Eq. (6), t = (v-v0)/a. Substituting this in Eq. 3.9a, we get,

This equation can also be obtained by substituting the value of t from Eq. (6) in to Eq. (8). Thus, we have obtained three important equations:

v = v0 + at

x = v0t + 1/2 a t²

v² = (v0)² + 2ax ……………………………………………..(11a)

connecting five quantities v0, v, a, t and x. These are kinematic equations of rectilinear motion for constant acceleration.

The set of Eq. (11a) were obtained by assuming that at t = 0, the position of the particle, x is 0. We can obtain a more general equation if we take the position coordinate at t = 0 as non-zero, say x0. Then Eqs. (11a) are modified (replacing x by x-x0 to:

v = v0 + at

x = x0t +v0t+ 1/2 a t² ………………………………….(11b)

v² = (v0)² + 2a(x-x0) …………………………………..(11c)

Example 3
Obtain equations of motion for constant acceleration using method of calculus.


By definition,

The advantage of this method is that it can be used for motion with non-uniform acceleration also.

Now, we shall use these equations to some important cases.

Example 4
A ball is thrown vertically upwards with a velocity of 20 ms-1 from the top of a mutistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground. (a) How high will the ball rise? and (b) how long will it be before the ball hits the ground? Take g = 10 ms-2.


(a) Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Fig. 13.

Now v0 = + 20 ms-1.

a = -g = -10 ms-2.

v = 0 ms-1

If the ball rises to height y from the point of launch, then using the equation

v² = (v0)² + 2a(y-y0)

we get

0 = (20)² + 2(-10)(y -y0)

Solving, we get, (y -y0) = 20 m.

(b) We can solve this part of the problem in two ways. Note carefully the two methods used.

FIRST METHOD: In the first method, we split the path in two parts: the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken t1 and t2. Since the velocity at B is zero, we have :

v = v0 + at

0 = 20 – 10 t1

Or, t1 =2 s

This is the time in going from A to B. From B or the point of the maximum height, the ball falls freely under the acceleration due to gravity. The ball is moving negative y direction. We use equation,

y = y0 + v0t + 1/2 a t²

We have y0 = 45 m, y = 0, v0 = 0 a = -g = –10 ms-2

0 = 45 +(1/2)(-10)(t2

Solving, we get, t2 = 3 s

Therefore, the total time taken by the ball before it hits the ground =t1 + t2 = 2 s + 3 s = 5 s.


The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation :

y = y0 + v0t + 1/2 a t²

We have y0 = 25 m, y = 0 m, v0 = 20 ms-1, a = -g = –10 ms-2

t = ?

0 = 25 +20 t + (1/2)(-10)

Solving, we get, t = 5 s

Note that the second method is better since we do not have to worry about the path of motion as the motion is under constant acceleration.

Example 5
Free-fall: Discuss the motion of an object under free fall. Neglect air resistance.


An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected the object said to be in free fall. If the height through which the object falls is small compared to earth’s radius, g can be taken to be constant, equal to 9.8 ms-2. Free fall is thus a case of motion with uniform acceleration.

We assume that the motion is in y-direction more correctly in – y-direction because we choose upward direction as positive. Since the acceleration due to gravity is always downward, it is in negative direction and we have,

a = -g = 9.8 ms-2

The object is released from rest at y = 0. Therefore, v0= 0 and the equations of motion become:

v = 0 – gt = 9.8 t ms-1

y = 0 – 1/2gt = 4.9 t2 m

v2 = 0 – gy = 19.6 y m2s-2

These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance. The variation of acceleration, velocity and distance with time have been plotted in Fig. 14(a), (b) and (c).


Fig. 14 Motion of an object under free fall (a) Variation of acceleration with time (b) Variation of velocity with time (c) Variation of distance with time.

Example 6
Galileo’s law of odd numbers: “The distance travelled, during equal intervals of time , by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, 1; 3; 5; 7………].” Prove it.


Let us divide the time interval of motion of an object under free fall in to many equal intervals ζ and find out the distances traversed during successive intervals of time. Since initial velocity is zero, we have

y 1/2 gt²

Using this, we can calculate the position of the object after different time intervals, 0, ζ, 2ζ, 3ζ ….. which are given in second column of Table 2. If we take (-1/2)gζ² as y0 – the position coordinate after first time interval ζ, then third column gives the positions in the unit of y0. The fourth column gives the distances traversed in successive ζs. We find that the distances are in simple ration 1:3:5:7:9:11 ….. as shown in the last column. This law was established by Galileo Galilei 91564-1642) who was the first to make quantitative studies of free fall.

Table 2

t y y in terms of
y0[m-(1/2) gζ² ]
Distance traversed in
successive intervals
Ratio of distances
0 0 0
ζ -(1/2)gζ² y0 y0 1
-4(1/2) gζ² 4y0 3y0 3
-9(1/2) gζ² 9y0 y0 5
-16(1/2) gζ² y0 y0 7
-25(1/2) gζ² y0 y0 9
-36(1/2) gζ² y0 y0 11

Example 7
Stopping distance of vehicles: When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (v0) and the braking capacity, or deceleration, -a that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of v0 and a.


Let the distance travelled by the vehicle before it stops be ds. Then, using equation of motion v² = v0² + 2ax, and noting that v = 0, we have the stopping distance. Thus the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).

For the car of a particular make, the braking distance was found to be 10 m, 20 m, 34 , and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s which are nearly consistent with the above formula. Stopping distance is an important factor considered in setting speed limits, for example is school zones.

Example 8
Reaction Time: When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on complexity of the situation and on an individual.

You can measure your reaction time by a simple experiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and fore finger (See Fig. 15 below). After you catch it, find the distance travelled by the ruler. In a particular case, d was found to bee 21.0 cm. Estimate reaction time.


The ruler drops under free fall. Therefore, v0 = 0, and g = -9.8 m s-². The distance travelled d and the reaction time tr are related by,

d = –1/2g r

Or, tr = √2d/g s

Given d = 21.0 cm and g =9.8 m s-² the reaction time is,

tr = √(2×0.21/9.8) s ≅ 0.2 s

Relative Velocity

You must be familiar with the experience of travelling in a train and being overtaken by another train in the same direction as you are. While the train must be travelling faster than you to be able to pass you, it does seem slower to you than it would be to someone standing on the ground and watching both the trains. In case both the trains have the same velocity with respect to the ground, then to you the other train would seem to be not moving at all. To understand such observations, we now introduce the concept of relative velocity.

Consider two objects A and B moving uniformly with average velocities vA and vB in one dimension, say along x-axis. (Unless otherwise specified, the velocities mentioned in this chapter are measured with reference to the ground). If xA (0) and xB (0) are positions of objects A and B, respectively at time t = 0, their positions xA (t) and xB (t) are given by:

xA (t) = xA (0) + vA (t) …………………………………………………………………(12a)

xB (t) = xB (0) + vB (t) …………………………………………………………………(12b)

Then, the displacement from object A to object B is given by,

xBA (t) = xB (t) + xA (t)

= [ xB (0) + xA (0)] + (vBvA )]t ……………………………………………………….(13)

Equation (13) is easily interpreted. It tells us that as seen from object A, object has a velocity vBvA in each unit of time. We say that the velocity of object B relative to object A is vBvA :

vBA = vB vA ……………………………………………………………………………..(14a)

Similarly, velocity of object A relative to object B is:

vAB= vA vB ………………………………………………………………………………(14b)

This shows: vBA = vB vA …………………………………………………………………(14c)

Now consider some special cases:

(a) If vB= vA , vB vA = 0. Then, fom Eq. (13), xB (t) – xA (t) = xB (0) – xA (0). Therefore, the two objects stay at a constant distance (xB (0) – xA (0)) apart, and their position-time graphs are straight lines parallel to each other as shown in Fig. 16. The relative velocity vAB or vBA is zero in this case.

(b)If vA ˃ vB , vB vA is negative. One graph is steeper than the other and they meet at a common point. For example, suppose vA = 20 m s-1 and xA (0) = 10 m; vB = 10 m s-1 and xB (0) = 40 m; then the at which they meet is t = 3s (Fig. 17). At this instant, they both are at a position xA (t) = xB (t) = 70 m. Thus object A overtakes object B at this time. In this case, vBA = 10 m s-1 – 20 m s-1 = -10 m s-1 = –vAB .

(c) Suppose vA , and VB are opposite signs. For example, if in the above example object A is moving with 20 m s-1 starting at xA (0) = 10 m and object B is moving with 20 m s-1 staring at xB (0) = 40 m, the two objects meet at t = 1s (Fig. 18). The velocity of B relative to A, vBA = [-10 – (20)] m s-1 = -30 m s-1 = – vAB. In this case, the magnitude of vBA or vAB (=30 m s-1) is greater than the magnitude of velocity of A or that of B. If the objects under consideration are two trains, then for a person sitting on either of the two, the other train seems to go very fast. Note that Eq. (14) are valid even if vA and vB represent instantaneous velocities.

Example 9
Two parallel rail tracks run north-south. Train A moves north with a speed of 54 km h-1, and train B moves south with a speed of 90 km h-1. What is the

(a) Velocity of B with respect to A?

(b) Velocity of ground with respect to B? and

(c) Velocity of a monkey running on the roof of the train A against its motion (with a velocity of 18 km h-1 with respect to train A) as observed by a man standing on the ground?


Choose the positive direction of x-axis to be from south to north. Then,

VA = +54 km h-1 = 15 m s-1

VB = – 90 km h-1 = – 25 m s-1

Relative velocity of B with respect to A = vB –vA = -40 m s-1. i.e., the train B appears to A to move with a speed of 40 m s-1 from north to south.

Relative velocity of ground with respect to B = 0 – vB = 25 m s-1

In (c), let the velocity of the monkey with respect to ground be vM . Relative velocity of the monkey with respect to A,

vMA = vM –vA = 18 km h-1 = -5 m s-1. Therefore, vMA= (15 -5) m s-1 = 10 m s-1.