Motion In A Straight Line
2.1 INTRODUCTION
Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves 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 twentyfour 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.
Motion is change in position of an object with time. How does the position change with time ? In this chapter, we shall learn how to describe motion. For this, we develop the concepts of velocity and acceleration. We shall confine ourselves to the study of motion of objects along a straight line, also known as rectilinear motion. For the case of rectilinear motion with uniform acceleration, a set of simple equations can be obtained. Finally, to understand the relative nature of motion, we introduce the concept of relative velocity.
In our discussions, we shall 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 reallife, the size of objects can be neglected and they can be considered as pointlike objects without much error. In Kinematics, we study ways to describe motion without going into the causes of motion. What causes motion described in this chapter and the next chapter forms the subject matter of Chapter 4.
2.2 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 ${\Delta T}$becomes infinitesimally small. In other words,
$\begin{aligned} v & =\lim _{\Delta \mathrm{t} \rightarrow 0} \frac{\Delta x}{\Delta t} \ & =\frac{\mathrm{d} x}{\mathrm{~d} t}\end{aligned}$
where the symbol lim ∆t→0 stands for the operation of taking limit as ∆tg0 of the quantity on its right. In the language of calculus, the quantity on the right hand side of Eq. (2.1a) is the differential coefficient of x with respect to t and is denoted by $\frac{\mathrm{d} x}{\mathrm{d} t}$ (see Appendix 2.1). It is the rate of change of position with respect to time, at that instant.
We can use Eq. (2.1a) for obtaining the value of velocity at an instant either graphically or numerically. Suppose that we want to obtain graphically the value of velocity at time t = 4 s (point P) for the motion of the car represented in Fig.2.1 calculation. Let us take ∆t = 2 s centred at t = 4 s. Then, by the definition of the average velocity, the slope of line $P_1P_2$ ( Fig. 2.1) gives the value of average velocity over the interval 3 s to 5 s
Fig. 2.1 Determining velocity from positiontime graph. Velocity at t = 4 s is the slope of the tangent to the graph at that instant.
Now, we decrease the value of $\Delta t$ from $2 \mathrm{~s}$ to 1 s. Then line $\mathrm{P}_1 \mathrm{P}_2$ becomes $\mathrm{Q}_1 \mathrm{Q}_2$ and its slope gives the value of the average velocity over the interval $3.5 \mathrm{~s}$ to $4.5 \mathrm{~s}$. In the limit $\Delta t \rightarrow 0$, the line $\mathrm{P}_1 \mathrm{P}_2$ becomes tangent to the positiontime curve at the point $\mathrm{P}$ and the velocity at $t$ $=4 \mathrm{~s}$ 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. 2.1, $x=0.08 t^3$. Table 2.1 gives the value of $\Delta x / \Delta t$ calculated for $\Delta t$ equal to $2.0 \mathrm{~s}$, $1.0 \mathrm{~s}, 0.5 \mathrm{~s}, 0.1 \mathrm{~s}$ and $0.01 \mathrm{~s}$ centred at $t=$ $4.0 \mathrm{~s}$. The second and third columns give the value of $t_1=\left(t\frac{\Delta t}{2}\right)$ and $t_2=\left(t+\frac{\Delta t}{2}\right)$ and the fourth and the fifth columns give the corresponding values of $x$, i.e. $x\left(t_1\right)=0.08 t_1^3$ and $x\left(t_2\right)=0.08 t_2^3$. The sixth column lists the difference $\Delta x=X\left(t_2\right)X\left(t_1\right)$ and the last column gives the ratio of $\Delta x$ and $\Delta t$, i.e. the average velocity corresponding to the value of $\Delta t$ listed in the first column.
Table 2.1 Limiting value of $\frac{\Delta x}{\Delta t}$ at $t=4 \mathrm{~s}$
(c)  to  $(6)$  $x(t)$  $(x)$  $\Delta:(m)$  $\Delta x / \Delta t$ $(a) \theta^y$ 

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  0.0384  3.8400 
We see from Table 2.1 that as we decrease the value of $\Delta t$ from $2.0 \mathrm{~s}$ to $0.010 \mathrm{~s}$, the value of the average velocity approaches the limiting value $3.84 \mathrm{~m} \mathrm{~s}^{1}$ which is the value of velocity at $t=4.0 \mathrm{~s}$, i.e. the value of $\frac{d x}{d t}$ at $t=4.0 \mathrm{~s}$.In this manner, we can calculate velocity at each instant for motion of the car.
The graphical method for the determination of the instantaneous velocity is always not a convenient method. For this, we must carefully plot the positiontime graph and calculate the value of average velocity as $\Delta t$ becomes smaller and smaller. It is easier to calculate the value of velocity at different instants if we have data of positions at different instants or exact expression for the position as a function of time. Then, we calculate $\Delta x / \Delta t$ from the data for decreasing the value of $\Delta t$ and find the limiting value as we have done in Table 2.1 or use differential calculus for the given expression and calculate $\frac{d x}{d t}$ at different instants as done in the following example.
Example 2.1 The position of an object moving along xaxis 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 = 0 s and t = 2.0 s. What is the average velocity between t = 2.0 s and t = 4.0 s ?
Answer In notation of differential calculus, the velocity is
$ v=\frac{d x}{d t}=\frac{d}{d t}\left(a+b t^2\right)=2 b t=5.0 t \mathrm{~m} \mathrm{~s}^{1} $
At $t=0 \mathrm{~s}, \quad V=0 \mathrm{~m} \mathrm{~s}^{1}$ and at $t=2.0 \mathrm{~s}$, $v=10 \mathrm{~m} \mathrm{~s}^{1}$.
$ \text { Average velocity }=\frac{x(4.0)x(2.0)}{4.02.0} $
$\begin{array}{r}=\frac{a+16 ba4 b}{2.0}=6.0 \times b \\ =6.0 \times 2.5=15 \mathrm{~m} \mathrm{~s}^{1}\end{array}$
Note that for uniform motion, velocity is the same as the average velocity at all instants
Instantaneous speed or simply speed is the magnitude of velocity. For example, a velocity of $+24.0 \mathrm{~m} \mathrm{~s}^{1}$ and a velocity of $24.0 \mathrm{~m} \mathrm{~s}^{1}$ both have an associated speed of $24.0 \mathrm{~m} \mathrm{~s}^{1}$. It should be noted that though average speed over a finite interval of time is greater or equal to the magnitude of the average velocity, instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant. Why so?
2.3 ACCELERATION
The velocity of an object, in general, changes during its course of motion. How to 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 could be described by the rate of change 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 the increasing distance of fall. This led to the concept of acceleration as the rate of change of velocity with time.
The average acceleration a over a time interval is defined as the change of velocity divided by the time interval :
$\bar{a}=\frac{v_2v_1}{t_2t_1}=\frac{\Delta v}{\Delta t}\quad \quad \quad \quad \quad (2.2)$
where $v_2$ and $v_1$ are the instantaneous velocities or simply velocities at time $t_2$ and $t_1$. It is the average change of velocity per unit time. The SI unit of acceleration is $\mathrm{m} \mathrm{s}^{2}$.
On a plot of velocity versus time, the average acceleration is the slope of the straight line connecting the points corresponding to $\left(v_2, t_2\right)$ and $\left(v_1, t_1\right)$.
Instantaneous acceleration is defined in the same way as the instantaneous velocity :
$ a=\lim\limits_{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{\mathrm{d} v}{\mathrm{~d} t} \quad \quad \quad \quad \quad (2.3) $
The acceleration at an instant is the slope of the tangent to the $vt$ curve at that instant.
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. Positiontime graphs for motion with positive, negative and zero acceleration are shown in Figs. 2.4 (a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration.
Fig. 2.2 Positiontime graph for motion with (a) positive acceleration; (b) negative acceleration, and (c) zero acceleration.
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 $V$ at $t$ $=0$ and $v$ at time $t$, we have
$ \bar{a}=\frac{vv_o}{t0} $
$\text { or, } v=v_o+a t \quad (2.4) $
Let us see how velocitytime graph looks like for some simple cases. Fig. 2.3 shows velocitytime graph for motion with constant acceleration for the following cases :
Fig. 2.3 Velocity–time graph for motions with constant acceleration. (a) Motion in positive direction with positive acceleration, (b) Motion in positive direction with negative acceleration, (c) Motion in negative direction with negative acceleration, (d) Motion of an object with negative acceleration that changes direction at time t1. Between times 0 to $t_1$, it moves in positive x  direction and between $t_1$ and $t_2$ it moves in the opposite direction.
(a) An object is moving in a positive direction with a positive acceleration.
(b) An object is moving in positive direction with a negative acceleration.
(c) An object is moving in negative direction with a negative acceleration.
(d) An object is moving in positive direction till time $t_1$, and then turns back with the same negative acceleration.
An interesting feature of a velocitytime 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 velocitytime graph is as shown in Fig. 2.4.
Fig. 2.4 Area under v–t curve equals displacement of the object over a given time interval.
The vt 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 coordinate axes, and you will arrive at the answer.
Note that the xt, vt, and at 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.
2.4 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 $\left(v_0\right)$, final velocity $(v)$ and acceleration (a). Equation (2.4) already obtained gives a relation between final and initial velocities $v$ and $v_0$ of an object moving with uniform acceleration $a$ :
$$ v=v_o+a t (2.4) $$
This relation is graphically represented in Fig. 2.5. The area under this curve is : Area between instants 0 and $t=$ Area of triangle $\mathrm{ABC}+$ Area of rectangle $\mathrm{OACD}$
$$ =\frac{1}{2}\left(vv_0\right) t+v_0 t $$
Fig. 2.5 Area under vt curve for an object with uniform acceleration.
As explained in the previous section, the area under vt curve represents the displacement. Therefore, the displacement x of the object is :
$$ x=\frac{1}{2}\left(vv_0\right) t+v_0 t \quad\quad \quad \quad \quad \quad (2.5) $$
But $\quad\quad \quad vv _0=a t$
Therefore, $\quad\quad x=\frac{1}{2} a t^2+v _0 t$
or, $\quad\quad \quad x=v_0 t+\frac{1}{2} a t^2 \quad (2.6)$
Equation (2.5) can also be written as
$$ \begin{align*} X & =\frac{V+v _{0}}{2} t \\ & =\bar{v} \cdot t \tag{2.7a} \\ \bar{v} & =\frac{v+v _{0}}{2} \text { (constant acceleration only) } \tag{2.7b} \end{align*} \quad $$
Equations (2.7a) and (2.7b) 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. (2.4), $t=\left(vv_0\right) / a$. Substituting this in Eq. (2.7a), we get
$$ \begin{align*} x & =\bar{v} t=\frac{v+v _{0}}{2} \cdot \frac{vv _{0}}{a}=\frac{v^{2}v _{0}^{2}}{2 a} \\ v^{2} & =v _{0}^{2}+2 a x \tag{2.8} \end{align*} $$
This equation can also be obtained by substituting the value of t from Eq. (2.4) into Eq. (2.6). Thus, we have obtained three important equations :connecting five quantities $v_0, v, a, t$ and $x$.
$$ \begin{gathered} v=v_0+a t \\ x=v_0 t+\frac{1}{2} a t^2 \\ v^2=v_0^2+2 a x \quad\quad\quad \quad (2.9a) \end{gathered} $$
These are kinematic equations of rectilinear motion for constant acceleration.
The set of Eq. (2.9a) 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 nonzero, say $x_0$. Then Eqs. (2.9a) are modified (replacing $x$ by $xx_0$ ) to :
$$ \begin{align*} v & =v _{0}+a t \\ x & =x _{0}+v _{0} t+\frac{1}{2} a t^{2} \tag{2.9b} \\ v^{2} & =v _{0}^{2}+2 a\left(xx _{0}\right) \tag{2.9c} \end{align*} $$
Example 2.2 Obtain equations of motion for constant acceleration using method of calculus.
Answer By definition
$$ \begin{aligned} & a=\frac{\mathrm{d} v}{\mathrm{~d} t} \\ & \mathrm{~d} v=a \mathrm{~d} t \end{aligned} $$
Integrating both sides
$$ \begin{aligned} \int_{v_0}^v \mathrm{~d} v & =\int_0^t a \mathrm{~d} t \\ & =a \int_0^t \mathrm{~d} t \quad \quad \quad \quad \text{( a is constant)}\\ vv_0 & =a t \\ v & =v_0+a t \end{aligned} $$
Further, $$ \begin{aligned} v & =\frac{\mathrm{d} x}{\mathrm{~d} t} \\ \mathrm{~d} x & =v \mathrm{~d} t \end{aligned} $$
Integrating both sides $$ \begin{aligned} \int_{x_0}^x \mathrm{~d} x=\int_0^t v \mathrm{~d} t & =\int_0^t\left(v_0+a t\right) \mathrm{d} t \\ xx_0 & =v_0 t+\frac{1}{2} a t^2 \\ x & =x_0+v_0 t+\frac{1}{2} a t^2 \end{aligned} $$
We can write
$$ \begin{aligned} & a=\frac{\mathrm{d} v}{\mathrm{~d} t}=\frac{\mathrm{d} v}{\mathrm{~d} x} \frac{\mathrm{d} x}{\mathrm{~d} t}=v \frac{\mathrm{d} v}{\mathrm{~d} x} \\ \\ & \text { or, } v \mathrm{~d} v=a \mathrm{~d} x \end{aligned} $$
Integrating both sides,
$$ \begin{aligned} & \int_{v_0}^v v \mathrm{~d} v=\int_{x_0}^x a \mathrm{~d} x \\ & \frac{v^2v_0^2}{2}=a\left(xx_0\right) \\ & v^2=v_0^2+2 a\left(xx_0\right) \end{aligned} $$
The advantage of this method is that it can be used for motion with nonuniform acceleration also.
Now, we shall use these equations to some important cases.
Example 2.3 A ball is thrown vertically upwards with a velocity of 20 m $s^{–1}$ from the top of a multistorey 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 m $s^{–2}$.
Answer (a) Let us take the $y$axis in the vertically upward direction with zero at the ground, as shown in Fig. 2.6.
$$ \begin{aligned} \text { Now } \quad v_o & =+20 \mathrm{~m} \mathrm{~s}^{1}, \\ a & =g=10 \mathrm{~m} \mathrm{~s}^{2}, \\ v & =0 \mathrm{~m} \mathrm{~s}^{1} \end{aligned} $$
If the ball rises to height $y$ from the point of launch, then using the equation $v^2=v_o^2+2 a\left(yy_0\right), 0=(20)^2+2(10)\left(yy_0\right),$
$$ \text{Solving, we get,} \left(yy_0\right)=20 \mathrm{~m}. $$
(b) We can solve this part of the problem in two ways.
Note carefully the methods used.
Fig. 2.6
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 $t_1$ and $t_2$ Since the velocity at B is zero, we have :
$$ \begin{aligned} & v=v_{\mathrm{o}}+a t \\ 0 & =2010 t_1 \\ \text { Or, } \quad \quad & t_1=2 \mathrm{~s} \end{aligned} $$
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 in negative $y$ direction. We use equation $$ y=y_0+v_0 t+\frac{1}{2} a t^2 $$
We have, $y_0=45 \mathrm{~m}, y=0, v_0=0, a=g=10 \mathrm{~m} \mathrm{~s}^{2}$
$$ 0=45+(1 / 2)(10) t_2^2 $$
Solving, we get $\mathrm{t}_2=3 \mathrm{~s}$ Therefore, the total time taken by the ball before it hits the ground $=t_1+t_2=2 \mathrm{~s}+3 \mathrm{~s}=5 \mathrm{~s}$.
SECOND METHOD : 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
$$ y=y_0+v_0 t+\frac{1}{2} a t^2 $$ $$ \begin{array}{ll} \text{Now} \quad \quad y_0=25 \mathrm{~m} & y=0 \mathrm{~m} \\ \quad \quad\quad \quad v_o=20 \mathrm{~m} \mathrm{~s}^{1}, & a=10 \mathrm{~m} \mathrm{~s}^{2}, \quad t=? \end{array} $$
$$ \begin{array}{ll} & 0=25+20 t+(1 / 2)(10) t^2 \\ \text { Or, } & \quad 5 t^220 t25=0 \end{array} $$
Solving this quadratic equation for $t$, we get
$$ t=5 \mathrm{~s} $$
Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration
Example 2.4 Freefall : Discuss the motion of an object under free fall. Neglect air resistance.
Answer 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 is said to be in free fall. If the height through which the object falls is small compared to the earth’s radius, g can be taken to be constant, equal to 9.8 m $s^{–2}$. Free fall is thus a case of motion with uniform acceleration
We assume that the motion is in ydirection, more correctly in –ydirection because we choose upward direction as positive. Since the acceleration due to gravity is always downward, it is in the negative direction and we have
$$ a=g=9.8 \mathrm{~m} \mathrm{~s}^{2} $$
The object is released from rest at $y=0$. Therefore, $v_0=0$ and the equations of motion become:
$$ \begin{array}{lll} v=0g t & =9.8 t &\mathrm{~m} \mathrm{~s}^{1} \\ y=01 / 2 g t^2 & =4.9 t^2 & \mathrm{~m} \\ v^2=02 g y & =19.6 y &\mathrm{~m}^2 \mathrm{~s}^{2} \end{array} $$
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. 2.7(a), (b) and (c).
(a)
(b)
(c)
Fig. 2.7 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 2.5 Galileo’s law of odd numbers : “The distances traversed, 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.
Answer Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distances traversed during successive intervals of time. Since initial velocity is zero, we have
$$ y=\frac{1}{2} g t^2 $$
Using this equation, we can calculate the position of the object after different time intervals, $0, \tau, 2 \tau, 3 \tau \ldots$ which are given in second column of Table 2.2. If we take $(1 / 2) g \tau^2$ as $y_0$  the position coordinate after first time interval $\tau$, then third column gives the positions in the unit of $y_0$. The fourth column gives the distances traversed in successive $\tau \mathrm{s}$. We find that the distances are in the simple ratio $1: 3: 5: 7: 9: 11 \ldots$ as shown in the last column. This law was established by Galileo Galilei (15641642) who was the first to make quantitative studies of free fall.
Example 2.6 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 $(v_0)$ 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 $v_o$ and a.
Table 2.2
$t$  $y$  $y \operatorname{mon}$ terme of $u_0\left[y_4\right) \in+1$ 
Ratho of dintereses 


0  0  0  
$=$  $(1 / 2) g \tau^2$  $y$  $y_0$  1 
$2 \tau$  $4(1 / 2) g \tau^2$  $4 y_n$  $y_0$ $3 y_0$ 
3 
$3 \tau$  $9(1 / 2) g \tau^2$  $9 y_0$  $5 y_9$  5 
$4 \tau$  $16(1 / 2) g \tau^2$  $16 y_0$  $7 y_0$  7 
$5 \tau$  $25(1 / 2) g \tau^2$  $25 y_0$  $9 y_0$  9 
$6 \tau$  $36(1 / 2) g \tau^2$  $36 y_0$  $11 y_0$  11 
Answer Let the distance travelled by the vehicle before it stops be $d_s$. Then, using equation of motion $v^2=v_0{ }^2+2 a x$, and noting that $v=0$, we have the stopping distance
$$ d_{\mathrm{s}}=\frac{v _0^2}{2 a} $$
Thus, the stopping distance is proportional to the square of the initial velocity. Doubling theinitial 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 m 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, in school zones.
Example 2.7 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 forefinger (Fig. 2.8). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be 21.0 cm. Estimate reaction time.
Fig. 2.8 Measuring the reaction time
Answer The ruler drops under free fall. Therefore, $v_o=0$, and $a=g=9.8 \mathrm{~m} \mathrm{~s}^{2}$. The distance travelled $d$ and the reaction time $t_r$ are related by
$$ d=\frac{1}{2} g t_r^2 $$
Or, $\quad t_r=\sqrt{\frac{2 d}{g}} \mathrm{~s}$
Given $d=21.0 \mathrm{~cm}$ and $g=9.8 \mathrm{~m} \mathrm{~s}^{2}$ the reaction time is
$$ t_r=\sqrt{\frac{2 \times 0.21}{9.8}} \mathrm{~s} \cong 0.2 \mathrm{~s} . $$
Summary
1. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval.
2. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval ∆t becomes infinitesimally small :
$$v=\lim _{\Delta t \rightarrow 0} \bar{v}=\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{\mathrm{d} x}{\mathrm{~d} t}$$
The velocity at a particular instant is equal to the slope of the tangent drawn on positiontime graph at that instant.
3. Average acceleration is the change in velocity divided by the time interval during which the change occurs :
$$ \bar{a}=\frac{\Delta v}{\Delta t} $$
4. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval $\Delta t$ goes to zero :
$$ a=\lim _{\Delta t \rightarrow 0} \bar{a}=\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{\mathrm{d} v}{\mathrm{~d} t} $$
The acceleration of an object at a particular time is the slope of the velocitytime graph at that instant of time. For uniform motion, acceleration is zero and the $x$  $t$ graph is a straight line inclined to the time axis and the $v$  $t$ graph is a straight line parallel to the time axis. For motion with uniform acceleration, $x$  $t$ graph is a parabola while the $v$  $t$ graph is a straight line inclined to the time axis.
5. The area under the velocitytime curve between times $t_1$ and $t_2$ is equal to the displacement of the object during that interval of time.
6. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement $x$, time taken $t$, initial velocity $v_0$, final velocity $v$ and acceleration $a$ are related by a set of simple equations called kinematic equations of motion :
$$ \begin{aligned} & v=v_o+a t \\ & x=v_0 t+\frac{1}{2} a t^2 \\ & v^2=v_0^2+2 a x \end{aligned} $$
if the position of the object at time $t=0$ is 0 . If the particle starts at $x=x_0, x$ in above equations is replaced by $\left(xx_0\right)$.
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Path length  $[\mathrm{L}]$  $\mathrm{m}$  
Displacement  $\Delta x$  $\mid L]$  $\mathrm{m}$  $=x_2x_1$ In one dimension, its sign indicates the direction. 
Velocity (a) Average 
$\bar{v}$  [LT*]  $\mathrm{ms}^2$  $=\frac{\Delta x}{\Delta t}$ 
(b) Instantaneous  $v$  $=\lim _{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{\mathrm{d} x}{\mathrm{~d} t}$ In one dimension, its sign indicates the direction. 
POINTS TO PONDER
1. The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration.
2. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the axis.
3. The sign of acceleration does not tell us whether the particle’s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed.
4. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have nonzero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity.
5. In the kinematic equations of motion [Eq. (2.9)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs.
6. The definitions of instantaneous velocity and acceleration (Eqs. (2.1) and (2.3)) are exact and are always correct while the kinematic equations (Eq. (2.9)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion.
Exercises
2.1 In which of the following examples of motion, can the body be considered approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
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Answer
Answer: (a), (b)
The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.
The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.
2.2 The positiontime (xt) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct entries in the brackets below ;
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice).
Fig. 2.9
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Answer
$\mathbf{A}$ lives closer to school than $\mathbf{B}$.
$\mathbf{A}$ starts from school earlier than $\mathbf{B}$.
$\mathbf{B}$ walks faster than $\mathbf{A}$.
$\mathbf{A}$ and $\mathbf{B}$ reach home at the same time.
$\mathbf{B}$ overtakes $\mathbf{A}$ once on the road.
Explanation:
In the given $xt$ graph, it can be observed that distance $OP<OQ$. Hence, the distance of school from the $\mathbf{A’s}$ home is less than that from $\mathbf{B’s}$ home.
In the given graph, it can be observed that for $x=0, t=0$ for $\mathbf{A}$, whereas for $x=0, t$ has some finite value for $\mathbf{B}$. Thus, A starts his journey from school earlier than $\mathbf{B}$.
In the given $xt$ graph, it can be observed that the slope of $\mathbf{B}$ is greater than that of $\mathbf{A}$. Since the slope of the $xt$ graph gives the speed, a greater slope means that the speed of $\mathbf{B}$ is greater than the speed $\mathbf{A}$.
It is clear from the given graph that both $\mathbf{A}$ and $\mathbf{B}$ reach their respective homes at the same time.
$\mathbf{B}$ moves later than A and his/her speed is greater than that of $\mathbf{A}$. From the graph, it is clear that $\mathbf{B}$ overtakes $\mathbf{A}$ only once on the road.
2.3 A woman starts from her home at $9.00 \mathrm{am}$, walks with a speed of $5 \mathrm{~km} \mathrm{~h}^{1}$ on a straight road up to her office $2.5 \mathrm{~km}$ away, stays at the office up to $5.00 \mathrm{pm}$, and returns home by an auto with a speed of $25 \mathrm{~km} \mathrm{~h}^{1}$. Choose suitable scales and plot the $x$  $t$ graph of her motion.
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Answer
Speed of the woman $=5 km / h$
Distance between her office and home $=2.5 km$
$ \begin{aligned} & \text{ Time taken }=\frac{\text{ Distance }}{\text{ Speed }} \\ & =\frac{2.5}{5}=0.5 h=30 \text{ min } \end{aligned} $
It is given that she covers the same distance in the evening by an auto.
Now, speed of the auto $=25 km / h$
$ \begin{aligned} & \text{ Time taken }=\frac{\text{ Distance }}{\text{ Speed }} \\ & =\frac{2.5}{25}=\frac{1}{10}=0.1 h=6 \text{ min } \end{aligned} $
The suitable $xt$ graph of the motion of the woman is shown in the given figure.
2.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is $1 \mathrm{~m}$ long and requires $1 \mathrm{~s}$. Plot the $x$  $t$ graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit $13 \mathrm{~m}$ away from the start.
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Answer
Distance covered with 1 step $=1 ~m$
Time taken $=1 ~s$
Time taken to move first $5 m$ forward $=5 ~s$
Time taken to move $3 ~m$ backward $=3 ~s$
Net distance covered $=53=2 ~m$
Net time taken to cover $2 ~m=8 ~s$
Drunkard covers $2 ~m$ in $8 ~s$.
Drunkard covered $4 ~m$ in $16 ~s$.
Drunkard covered $6 ~m$ in $24 ~s$.
Drunkard covered $8 ~m$ in $32 ~s$.
In the next $5 s$, the drunkard will cover a distance of $5 ~m$ and a total distance of $13 ~m$ and falls into the pit.
Net time taken by the drunkard to cover $13 ~m=32+5=37 ~s$
The $xt$ graph of the drunkard’s motion can be shown as:
2.5 A car moving along a straight highway with speed of $126 \mathrm{~km} \mathrm{~h}^{1}$ is brought to a stop within a distance of $200 \mathrm{~m}$. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
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Answer
Initial velocity of the car, $u=126 km / h=35 m / s$
Final velocity of the car, $v=0$
Distance covered by the car before coming to rest, $s=200 m$
Retardation produced in the car $=a$
From third equation of motion, $a$ can be calculated as:
$v^{2}u^{2}=2 a s$
$(0)^{2}(35)^{2}=2 \times a \times 200$
$a=\frac{35 \times 35}{2 \times 200}=3.06 m / s^{2}$
From first equation of motion, time $(t)$ taken by the car to stop can be obtained as:
$v=u+a t$
$t=\frac{vu}{a}=\frac{35}{3.06}=11.44 s$
2.6 A player throws a ball upwards with an initial speed of $29.4 \mathrm{~m} \mathrm{~s}^{1}$.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the $x=0 \mathrm{~m}$ and $t=0 \mathrm{~s}$ to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of $x$axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands ? (Take $g=9.8 \mathrm{~m} \mathrm{~s}^{2}$ and neglect air resistance).
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Answer
Downward
Velocity $=0$, acceleration $=9.8 ~m / s^{2}$
$x>0$ for both up and down motions, $v<0$ for up and $v>0$ for down motion, $a>0$ throughout the motion
$44.1 ~m, 6 ~s$
Explanation:
Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.
At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., $9.8 ~m / s^{2}$.
During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.
Initial velocity of the ball, $u=29.4 ~m / s$
Final velocity of the ball, $v=0$ (At maximum height, the velocity of the ball becomes zero)
Acceleration, $a=g=9.8 ~m / s^{2}$
From third equation of motion, height (s) can be calculated as:
$ \begin{aligned} & v^{2}u^{2}=2 g s \\ & s=\frac{v^{2}u^{2}}{2 g} \\ \\ & =\frac{(0)^{2}(29.4)^{2}}{2 \times(9.8)}=44.1 m \end{aligned} $
From first equation of motion, time of ascent $(t)$ is given as: $v=u+a t$
$t=\frac{vu}{a}=\frac{29.4}{9.8}=3 s$
Time of ascent $=$ Time of descent
Hence, the total time taken by the ball to return to the player’s hands $=3+3=6 s$.
2.7 Read each statement below carefully and state with reasons and examples, if it is true or false ;
A particle in onedimensional motion
(a) with zero speed at an instant may have nonzero acceleration at that instant
(b) with zero speed may have nonzero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Show Answer
Answer
True
False
True
False
Explanation:
When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.
Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.
A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero.
This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards.
This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.
2.8 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speedtime graph of its motion between t = 0 to 12 s.
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Answer
Ball is dropped from a height, $s=90 ~m$
Initial velocity of the ball, $u=0$
Acceleration, $a=g=9.8 ~m / s^{2}$
Final velocity of the ball $=v$
From second equation of motion, time $(t)$ taken by the ball to hit the ground can be obtained as:
$ \begin{aligned} & s=u t+\frac{1}{2} a t^{2} \\ \\ & 90=0+\frac{1}{2} \times 9.8 t^{2} \\ \\ & t=\sqrt{18.38}=4.29 ~s \end{aligned} $
From first equation of motion, final velocity is given as:
$v=u+a t$
$=0+9.8 \times 4.29=42.04 ~m / s$
Rebound velocity of the ball, $u_r=\frac{9}{10} v=\frac{9}{10} \times 42.04=37.84 ~m / s$
Time $(t)$ taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
$v=u_r+a t^{\prime}$
$0=37.84+(9.8) t^{\prime}$
$t^{\prime}=\frac{37.84}{9.8}=3.86 s$
Total time taken by the ball $=t+t^{\prime}=4.29+3.86=8.15 s$
As the time of ascent is equal to the time of descent, the ball takes $3.86 s$ to strike back on the floor for the second time.
The velocity with which the ball rebounds from the floor $=\frac{9}{10} \times 37.84=34.05 ~m / s$
Total time taken by the ball for second rebound $=8.15+3.86=12.01 ~s$
The speedtime graph of the ball is represented in the given figure as:
2.9 Explain clearly, with examples, the distinction between :
(a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider onedimensional motion only].
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Answer
The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle.
The total path length of a particle is the actual path length covered by the particle in a given interval of time.
For example, suppose a particle moves from point $A$ to point $B$ and then, comes back to a point, $C$ taking a total time $t$, as shown below. Then, the magnitude of displacement of the particle $=AC$.
Whereas, total path length $=AB+BC$
It is also important to note that the magnitude of displacement can never be greater than the total path length. However, in some cases, both quantities are equal to each other.
(b)
Magnitude of average velocity $=\frac{\text{ Magnitude of displacement }}{\text{ Time interval }}$
For the given particle,
Average velocity $=\frac{AC}{t}$
$ \begin{aligned} \text{ Average speed } & =\frac{\text{ Total path length }}{\text{ Time interval }} \\ \\ & =\frac{AB+BC}{t} \end{aligned} $
Since $(A B+B C)>A C$, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.
2.10 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km $h^{–1.}$ Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ?[ Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Fig. 2.10
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Answer
Time taken by the man to reach the market from home, $t_1=\frac{2.5}{5}=\frac{1}{2} h=30 min$
Time taken by the man to reach home from the market, $t_2=\frac{2.5}{7.5}=\frac{1}{3} h=20 min$
Total time taken in the whole journey $=30+20=50 min$
$$ \begin{align*} & \text{ Average velocity }=\frac{\text{ Displacement }}{\text{ Time }}=\frac{2.5}{\frac{1}{2}}=5 km / h \tag{a(i)} \end{align*} $$
$$ \begin{align*} & \text{ Average speed }=\frac{\text{ Distance }}{\text{ Time }}=\frac{2.5}{\frac{1}{2}}=5 km / h \tag{a(ii)} \end{align*} $$
Time $=50 \min =\frac{5}{6} h$
Net displacement $=0$
Total distance $=2.5+2.5=5 km$
$$ \begin{align*} & \text{ Average velocity }=\frac{\text{ Displacement }}{\text{ Time }}=0 \tag{b(i)} \end{align*} $$
$$ \begin{align*} & \text{ Average speed }=\frac{\text{ Distance }}{\text{ Time }}=\frac{5}{(\frac{5}{6})}=6 km / h \quad \ldots \tag{b(ii)} \end{align*} $$
Speed of the man $=7.5 km$
Distance travelled in first $30 min=2.5 km$
Distance travelled by the man (from market to home) in the next $10 min$
$ =7.5 \times \frac{10}{60}=1.25 km $
Net displacement $=2.51.25=1.25 km$
Total distance travelled $=2.5+1.25=3.75 km$
$ \begin{aligned} & \text{ Average velocity }=\frac{1.25}{(\frac{40}{60})}=\frac{1.25 \times 3}{2}=1.875 km / h \quad \ldots(a(iii)) \\ & \text{ Average speed }=\frac{3.75}{(\frac{40}{60})}=5.625 km / h \quad \ldots(b(iii)) \end{aligned} $
2.11 In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
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Answer
Instantaneous velocity is given by the first derivative of distance with respect to time i.e.,
$ v_{ln}=\frac{d x}{d t} $
Here, the time interval $d t$ is so small that it is assumed that the particle does not change its direction of motion. As a result, both the total path length and magnitude of displacement become equal is this interval of time.
Therefore, instantaneous speed is always equal to instantaneous velocity.
2.12 Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent onedimensional motion of a particle.
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Answer
The given $x$  $t$ graph, shown in (a), does not represent onedimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.
The given $vt$ graph, shown in (b), does not represent onedimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
The given $vt$ graph, shown in (c), does not represent onedimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.
The given $vt$ graph, shown in (d), does not represent onedimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.
2.13 Figure 2.11 shows the $xt$ plot of onedimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph.
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Answer
No
The $xt$ graph of a particle moving in a straight line for $t<0$ and on a parabolic path for $t$ $>0$ cannot be shown as the given graph. This is because, the given particle does not follow the trajectory of path followed by the particle as $t=0, x=0$. A physical situation that resembles the above graph is of a freely falling body held for sometime at a height
2.14 A police van moving on a highway with a speed of 30 km $h^{–1}$ fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km $h^{–1}$. If the muzzle speed of the bullet is 150 $ms^{–1}$, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car).
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Answer
Speed of the police van, $v_p=30 km / h=8.33 m / s$
Muzzle speed of the bullet, $v_b=150 m / s$
Speed of the thief’s car, $v_t=192 km / h=53.33 m / s$
Since the bullet is fired from a moving van, its resultant speed can be obtained as:
$=150+8.33=158.33 m / s$
Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as:
$v _{b t}=v_bv_t$
$=158.3353.33=105 m / s$
2.15 Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
Fig. 2.12
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Answer (a)The given $x$  $t$ graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.
(b)In the given $v$tgraph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.
(c)The given $a$t graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.
2.16 Figure 2.13 gives the $xt$ plot of a particle executing onedimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
Fig. 2.13
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Answer
Negative, Negative, Positive (at $t=0.3 s$ )
Positive, Positive, Negative (at $t=1.2 s$ )
Negative, Positive, Positive (at $t=1.2 s$)
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:
$a=\omega^{2} x \omega \to$ angular frequency.
$t=0.3 s$
In this time interval, $x$ is negative. Thus, the slope of the $xt$ plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.
$t=1.2 s$
In this time interval, $x$ is positive. Thus, the slope of the $xt$ plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative. $t=1.2 s$
In this time interval, $x$ is negative. Thus, the slope of the $xt$ plot will also be negative. Since both $x$ and $t$ are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.
2.17 Figure 2.14 gives the $xt$ plot of a particle in onedimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.
Fig. 2.14
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Answer
Interval 3 (Greatest), Interval 2 (Least)
Positive (Intervals $1 \& 2$ ), Negative (Interval 3)
The average speed of a particle shown in the $xt$ graph is obtained from the slope of the graph in a particular interval of time.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.
2.18 Figure 2.15 gives a speedtime graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?
Fig. 2.15
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Answer
Average acceleration is greatest in interval 2
Average speed is greatest in interval 3
$v$ is positive in intervals 1,2 , and 3
$a$ is positive in intervals 1 and 3 and negative in interval 2
$a=0$ at $A, B, C, D$
Acceleration is given by the slope of the speedtime graph. In the given case, it is given by the slope of the speedtime graph within the given interval of time.
Since the slope of the given speedtime graph is maximum in interval 2, average acceleration will be the greatest in this interval.
Height of the curve from the timeaxis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.
In interval 1:
The slope of the speedtime graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.
In interval 2:
The slope of the speedtime graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.
In interval 3:
The slope of the speedtime graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.
Points A, B, C, and D are all parallel to the timeaxis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.