Motion in a Straight Line: Class 11 Physics · Chapter 2

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

70 min
27 practice questions
17 MCQs

What you will learn in Motion in a Straight Line

Describe position, displacement, path length, average velocity and average speed for motion along a line, with a consistent sign convention
Find instantaneous velocity and acceleration as limits, as slopes of x–t and v–t graphs, and by differentiation
Read direction of motion and speeding up or slowing down from graphs, and get displacement from the area under a v–t graph
Derive the kinematic equations for constant acceleration by the area method and by integration
Solve problems on vertical throws, free fall, stopping distance and reaction time
Use relative velocity to find when and where two objects moving along a line meet or overtake

Original companion text aligned to NCERT Physics Class 11, Chapter 2, pp. 13–26. Not affiliated with or endorsed by NCERT.

2.1 Introduction

A cricket ball leaving the bowler's hand, a lift rising through a tower, an autorickshaw braking at a red light: in each, something keeps changing its position. This chapter builds the language physicists use to say exactly how the position changes. Three ideas do the job, position, velocity and acceleration, and we will make each one precise enough to calculate with.

The plan has three stages. We first pin down velocity and acceleration exactly, with graphs and with calculus. Then, for the common special case of a fixed acceleration, we derive three short equations that predict where a body will be and how fast it will be moving. Finally we ask how a motion looks to an observer who is moving too. None of this asks why a body speeds up; forces wait for Chapter 4, Laws of Motion. Pure description of motion is called kinematics (शुद्धगतिकी).

Two simplifications run through the chapter. The motion stays on one straight line, like a train on a straight stretch of track or a stone dropped down a well. This is rectilinear motion (सरल रेखीय गति), and a single number $x$ fixes the position. And each moving body shrinks to a point object: we forget its size and shape and follow one point on it, which is safe when the body is tiny compared with the ground it covers.

Position, displacement and average velocity: a recap

You met these ideas in Class 9; now they get signs and symbols. To give a position we first fix a reference point, the origin $O$ , and a positive direction along the line. Points on the positive side have positive $x$ , points on the other side negative $x$ . The choice is ours, but it must be stated before any sign is written, and kept for the whole problem.

Suppose an object is at $x_{1}$ at time $t_{1}$ and at $x_{2}$ at a later time $t_{2}$ . Its displacement (विस्थापन) over the interval is $Δ x = x_{2} - x_{1}$ : positive for a net shift in the positive direction, negative for a net shift the other way. The path length is the total distance actually covered along the route. The two differ as soon as the object turns back (Fig. 2.1).

3/3Displacement $= x_{R} - x_{P} = + 4 k m$ , but the path length is $8 k m$ .

Fig. 2.1Path length counts every kilometre driven; displacement only compares where the trip ends with where it began.

A horizontal number line in kilometres from minus 2 to 6 with the origin O at the tea stall and the positive direction to the right. Step 1 shows the axis and the starting point P at minus 1 kilometre. Step 2 adds an arrow from P to Q at plus 5, labelled 6 kilometres out, and a second arrow back from Q to R at plus 3, labelled 2 kilometres back. Step 3 draws the displacement arrow from P straight to R, labelled displacement plus 4 kilometres, and notes that the path length is 8 kilometres.

Dividing by the time taken gives two different averages. The average velocity (औसत वेग) uses the displacement, and the average speed (औसत चाल) uses the path length:

Average velocity

\overset{ˉ}{v} = \frac{x_{2} - x_{1}}{t_{2} - t_{1}} = \frac{Δ x}{Δ t}

Average speed

average speed = \frac{path length}{Δ t}

The path length can never be less than the size of the displacement, so the average speed is never less than the magnitude of the average velocity. They are equal only if the object does not reverse during the interval. On a position–time ( $x$ – $t$ ) graph, $\overset{ˉ}{v}$ is the slope of the straight chord joining the two points. In uniform motion the graph is itself a straight line, so every chord has the same slope: the velocity is the same over every interval.

Example

An e-rickshaw's errand

Take the tea stall as the origin and the road towards the school as positive. An e-rickshaw starts at $P$ , $1.0 k m$ on the far side of the stall, drives $6.0 k m$ to $Q$ in 15 min, then turns and drives $2.0 k m$ back to $R$ in 5 min. Find its displacement, path length, average velocity and average speed for the whole trip.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

For which question can the moving body be treated as a point object?

How fast the tip of a spinning fan blade movesHow a gymnast's body turns in a somersaultHow long a satellite takes to orbit the EarthHow the edge of a door moves as it swings shut

Show answer

Answer (C) How long a satellite takes to orbit the Earth

A satellite a few metres across, on an orbit thousands of kilometres in radius, is effectively a point. In the other three the question is about how different parts of the body move, so its size and shape matter.

2.2 Instantaneous velocity and speed

An average velocity hides detail. A car that covers $60 k m$ in an hour may have crawled through traffic and then sped along a highway; the average tells you nothing about its speedometer reading at 10:15. To capture the velocity at an instant, shrink the interval. Work out the average velocity over a short interval $Δ t$ starting at time $t$ , then make $Δ t$ smaller and smaller and see what value the average settles on.

That limiting value is the instantaneous velocity (तात्क्षणिक वेग) at time $t$ ; when physicists say 'velocity' without a qualifier, this is what they mean. In calculus notation it is the derivative $d x / d t$ :

Instantaneous velocity

v = \lim_{Δ t \to 0} \frac{Δ x}{Δ t} = \frac{d x}{d t}

(2.1)

where

$Δ x$: the displacement in the short interval $Δ t$
$d x / d t$: the rate of change of position at the instant $t$

There are three ways to find this limit. Graphically: as the second point of a chord on the $x$ – $t$ graph slides towards the first, the chord turns into the tangent at the first point (Fig. 2.2). So the tangent's slope at a point of the $x$ – $t$ graph gives the velocity at that moment.

4/4In the limit the chord becomes the tangent at $P$ : $v = 6.0 m / s$ .

Fig. 2.2As

Δ t

shrinks, the chord from

P

turns into the tangent at

P

; its slope is the velocity at

t = 2 s

.NCERT Fig. 2.1 · p. 14

A position-time graph, position from 0 to 16 metres against time from 0 to 3.5 seconds, shows the rising curve x equals 1.5 t squared, with point P at 2 seconds and 6 metres. Step 1 draws the chord from P to the point at 3 seconds, slope 7.5 metres per second. Step 2 moves the second point to 2.5 seconds, slope 6.75. Step 3 moves it to 2.1 seconds, slope 6.15, almost touching the curve. Step 4 shows the tangent at P with slope 6.0 metres per second, the earlier chords faded behind it.

Numerically: compute $Δ x / Δ t$ for smaller and smaller $Δ t$ . A motorbike pulling away from a traffic signal has $x = 1.5 t^{2}$ (metres, with $t$ in seconds); Table 2.1 finds its velocity at $t = 2.0 s$ . Intervals taken forward from 2 s and backward from 2 s close in on the same value, $6.0 m / s$ , one from above and one from below.

Table 2.1Average velocity of the motorbike (

x = 1.5 t^{2}

) over shrinking intervals at

t = 2.0 s

NCERT Table 2.1 · p. 14

$Δ t$ (s)	forward $Δ x / Δ t$ (m/s)	backward $Δ x / Δ t$ (m/s)
1.0	7.5	4.5
0.5	6.75	5.25
0.1	6.15	5.85
0.01	6.015	5.985
0.001	6.0015	5.9985

With calculus: when $x (t)$ is a formula, differentiate it. For $x = 1.5 t^{2}$ , $v = d x / d t = 3.0 t$ , which gives $6.0 m / s$ at $t = 2.0 s$ , exactly the number the table was homing in on. Calculus is usually quickest; the table shows what a derivative means. In uniform motion every chord already lies along the straight $x$ – $t$ line, so the velocity at any instant equals the average velocity over any interval.

Example 2.1NCERT Example 2.1 · p. 15

Velocity of a bead on a wire

A bead slides along a straight wire with position $x = 6.0 t - 0.50 t^{3}$ , where $x$ is in metres and $t$ in seconds. Find (a) its velocity at $t = 1.0 s$ and at $t = 2.0 s$ ; (b) its average velocity between $t = 0$ and $t = 2.0 s$ .

Try it yourself first. Then reveal the solution one step at a time.

Two autorickshaws pass each other at a crossing, one heading north and one heading south, both at $8 m / s$ . With north as positive their velocities are $+ 8$ and $- 8 m / s$ , yet their speedometers agree. A speedometer shows the speed (चाल), the magnitude $∣ v ∣$ of the velocity, which is blind to direction.

Quick check

1 of 3

How do you read the velocity at a particular moment off an

x

–

t

graph?

Take the slope of the tangent thereRead the height of the graph thereFind the area under the graph up to thereTake the slope of the chord from the origin

Show answer

Answer (A) Take the slope of the tangent there

Δ t \to 0

the chord becomes the tangent, so its slope gives

v

. The chord from the origin gives the average velocity since

t = 0

, not the velocity at the instant.

2.3 Acceleration

Velocity usually changes too: a bus pulls away from a stop, a stone thrown upward slows down. Should we measure the change per second or per metre? Galileo himself first picked the wrong answer (see the box). Watch a stone dropped from rest: after 1 s it moves at $9.8 m / s$ , after 2 s at $19.6 m / s$ , the same gain in every second. Per metre the picture is messy: the first $4.9 m$ of fall add $9.8 m / s$ , but the next $14.7 m$ add only another $9.8 m / s$ . A rate that stays constant is far more useful, so acceleration is defined per unit time.

The acceleration (त्वरण) measures this rate. If the velocity changes from $v_{1}$ at time $t_{1}$ to $v_{2}$ at time $t_{2}$ , the average acceleration over the interval is

Average acceleration

\overset{ˉ}{a} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}} = \frac{Δ v}{Δ t}

(2.2)

Its unit is $(m / s) / s = m s^{- 2}$ . On a velocity–time ( $v$ – $t$ ) graph, $\overset{ˉ}{a}$ is the slope of a chord, just as $\overset{ˉ}{v}$ was a chord slope on the $x$ – $t$ graph. Let the chord shrink to a tangent and you get the instantaneous acceleration:

Instantaneous acceleration

a = \lim_{Δ t \to 0} \frac{Δ v}{Δ t} = \frac{d v}{d t}

(2.3)

On a line, 'direction' has only two values, $+$ and $-$ , so $a$ is simply a signed number. (In Chapter 3, on motion in a plane, a velocity can swing round without changing its size, and that turning also counts as acceleration.)

Example

At rest, yet accelerating

For the bead of Example 2.1, find (a) its average acceleration from $t = 0$ to $t = 2.0 s$ ; (b) its acceleration at $t = 2.0 s$ , the instant it is at rest.

Try it yourself first. Then reveal the solution one step at a time.

The $x$ – $t$ graph also reveals the sign of the acceleration, through its curvature. The velocity is the slope of the $x$ – $t$ graph. With positive acceleration that slope keeps increasing, so the graph bends upward. With negative acceleration the slope keeps decreasing and the graph bends downward. With zero acceleration the slope is fixed and the graph is straight (Fig. 2.3).

1/3Slope rising: the graph bends upward, $a > 0$ .

Fig. 2.3The curvature of an

x

–

t

graph shows the sign of the acceleration.NCERT Fig. 2.2 · p. 16

A position-time axis with one curve per step and three short tangent segments drawn on it. Step 1: the curve bends upward and the tangents get steeper from left to right, labelled slope keeps rising, for positive acceleration. Step 2: the curve bends downward and the tangents get flatter, labelled slope keeps falling, for negative acceleration. Step 3: a straight line whose three tangents all lie along it with the same slope, for zero acceleration.

From here on the acceleration is held fixed, as for a stone in flight or a train pulling out with steady thrust. With constant acceleration every chord and every tangent of the $v$ – $t$ graph has the same slope $a$ . If the velocity is $v_{0}$ when the clock starts ( $t = 0$ ), then after a time $t$ it has grown by $a t$ :

Velocity under constant acceleration

v = v_{0} + a t

(2.4)

So under constant acceleration the $v$ – $t$ graph is straight, with slope $a$ and intercept $v_{0}$ . Fig. 2.4 shows the four kinds of line you will meet. Ask two questions of each: which side of the time axis is the line on? (that is the direction of motion) and is the line moving away from the axis or towards it? (speeding up or slowing down).

4/4 $a < 0$ throughout: forward, momentarily at rest where the line crosses the axis, then backward.

Fig. 2.4Above the axis means moving in the

+ x

direction; a line heading away from the axis means speeding up.NCERT Fig. 2.3 · p. 16

A velocity-time graph with plus values above and minus values below the time axis, and above it a road with a car carrying a blue velocity arrow and a vermillion acceleration arrow. Step 1: a line rising from plus 1 to plus 4, car moving right with both arrows pointing right, speeding up. Step 2: a line falling from plus 4 to plus 1, car moving right, acceleration arrow pointing left, slowing down. Step 3: a line falling from minus 1 to minus 4, car facing left, both arrows pointing left, speeding up. Step 4: a line from plus 3 down through zero to minus 3; the acceleration points left, the velocity points right early and left later, so the car stops and turns back where the line crosses the axis.

Over a short slice of time $Δ t$ the velocity barely changes, so the object moves about $v Δ t$ . That product is exactly the area of a thin strip, of height $v$ and width $Δ t$ , under the $v$ – $t$ graph. Stack such strips across an interval and their areas add up to the whole displacement. Make them thinner and the staircase hugs the curve; in the limit the sum becomes an integral (Fig. 2.5). So: the area between a $v$ – $t$ graph and the time axis over an interval equals the displacement in that interval. With constant velocity no slicing is needed: a car holding $4 m / s$ for $5 s$ goes $20 m$ , the area of one rectangle.

3/3Area above the axis is $+ 4.5 m$ , below it $- 2.0 m$ : displacement $2.5 m$ , path length $6.5 m$ .

Fig. 2.5Area under a

v

–

t

graph is displacement; area below the axis counts as negative.NCERT Fig. 2.4 · p. 17

Velocity in metres per second against time in seconds. Step 1: a horizontal line at 4 metres per second for 5 seconds with the rectangle under it shaded and labelled area equals 20 metres. Step 2: a rising curve with ten thin rectangular strips under it, labelled thin strips. Step 3: a straight line falling from 3 metres per second through zero at 3 seconds to minus 2 at 5 seconds; the triangle above the axis is labelled plus 4.5 metres and the dashed triangle below it minus 2.0 metres, giving displacement 2.5 metres and path length 6.5 metres.

Area below the time axis comes from negative velocity, so it counts as negative displacement. To get the path length instead, add the sizes of all the areas.

Quick check

1 of 4

With up as positive, a skydiver who has just opened her parachute is still moving down but slowing. The signs of

v

and

a

are

v < 0

and

a < 0

v > 0

and

a < 0

v < 0

and

a > 0

v > 0

and

a > 0

Show answer

Answer (C) $v < 0$ and $a > 0$

Moving down gives

v < 0

. Slowing down means

a

has the opposite sign to

v

, so

a > 0

: an upward acceleration. With

a < 0

(option A) she would be speeding up.

2.4 Kinematic equations for uniformly accelerated motion

Equation (2.4) answers 'how fast?' for uniformly accelerated motion. 'How far?' comes from the area under the straight $v$ – $t$ line. Start the clock ( $t = 0$ ) with the object at the origin, so that its position $x$ at time $t$ is also its displacement. Its velocity is $v_{0}$ at the start and $v$ at time $t$ .

DerivationThe equations from the area under the v–t line

In Fig. 2.6 the line rises from $v_{0}$ at time 0 to $v$ at time $t$ . Split the area under it into a rectangle of height $v_{0}$ and, on top of it, a triangle of height $v - v_{0}$ . The area, and so the displacement, is

2/3Triangle: base $t$ , height $v - v_{0} = a t$ , area $\frac{1}{2} a t^{2}$ .

Fig. 2.6Rectangle plus triangle gives

x = v_{0} t + \frac{1}{2} a t^{2}

; the dashed line gives

x = \overset{ˉ}{v} t

.NCERT Fig. 2.5 · p. 17

Velocity against time; a straight line rises from v-nought at time zero to v at time t. Step 1 shades the rectangle of height v-nought and width t, labelled v-nought t. Step 2 adds the triangle on top, of height v minus v-nought, labelled one half a t squared. Step 3 draws a dashed horizontal line at the average of v-nought and v; the small triangle cut off above it on the right exactly fills the gap below it on the left, so the area equals the average velocity times t.

x = v_{0} t + \frac{1}{2} (v - v_{0}) t

(2.5)

From (2.4), $v - v_{0} = a t$ . Substituting,

x = v_{0} t + \frac{1}{2} a t^{2}

(2.6)

Equation (2.5) also regroups as $x = \frac{1}{2} (v_{0} + v) t$ . The object covers the same ground as if it had moved steadily at the mean of its starting and final velocities. In the last step of the figure, the dashed line at that height cuts off a triangle above it that exactly fills the gap below it.

x = \overset{ˉ}{v} t

(2.7a)

constant acceleration only

\overset{ˉ}{v} = \frac{v_{0} + v}{2}

(2.7b)

A time-free equation follows by multiplying two results we already have: (2.4) gives $v - v_{0} = a t$ , and (2.7a) with (2.7b) gives $v + v_{0} = 2 x / t$ . In the product the $t$ 's cancel:

(v - v_{0}) (v + v_{0}) = a t \cdot \frac{2 x}{t}

v^{2} = v_{0}^{2} + 2 a x

(2.8)

Together these connect $v_{0}$ , $v$ , $a$ , $t$ and $x$ ; know any three and the other two follow:

Kinematic equations

\begin{aligned} v & = v_{0} + a t \\ x & = v_{0} t + \frac{1}{2} a t^{2} \\ v^{2} & = v_{0}^{2} + 2 a x \end{aligned}

(2.9a)

If the object starts at $x_{0}$ rather than at the origin, its displacement is $x - x_{0}$ . The first equation is unchanged and the other two become

x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}

(2.9b)

v^{2} = v_{0}^{2} + 2 a (x - x_{0})

(2.9c)

InteractiveMotion lab: x–t, v–t and a–t together

An interactive kinematics lab. Four presets (cruise, speed up, thrown up, backward start) or three sliders set the initial position from minus 20 to 20 metres, the initial velocity from minus 20 to 20 metres per second and a constant acceleration from minus 10 to 10 metres per second squared. Pressing Play runs 4 seconds of motion: a dot moves along a horizontal x-axis while position-time, velocity-time and acceleration-time graphs grow up to the current time. Area under the velocity graph is shaded, green above the time axis and vermillion with dashes below it. A time slider stops at any instant. A readout gives x, v and a, whether the object is speeding up or slowing down, the displacement and path length so far, and the average velocity and average speed.

Fig. 2.7Set

x_{0}

v_{0}

and

a

, press Play, and compare the three graphs. Start with the Thrown up preset.

In the lab, give the object a positive $v_{0}$ and a negative $a$ and watch it stop and come back. The $v$ – $t$ line crosses the time axis at the turning point, the $x$ – $t$ curve peaks at that same instant, and from then on the path length pulls ahead of the displacement.

Example 2.2NCERT Example 2.2 · p. 18

The equations by calculus

Derive the kinematic equations by integration for an object with constant acceleration $a$ that has velocity $v_{0}$ and position $x_{0}$ at $t = 0$ . Then show why this method is more general, using an acceleration that grows with time, $a = 2 t$ (SI units), with $v_{0} = 0$ and $x_{0} = 0$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 2.3NCERT Example 2.3 · p. 18

A stone thrown up from a cliff

From the edge of a cliff $20 m$ above a river, a student throws a stone straight up at $15 m / s$ . It rises, falls past the cliff edge and drops into the river. Take $g = 10 m / s^{2}$ and ignore air resistance. Find (a) the greatest height of the stone above the river, (b) when it hits the water and (c) how fast it is moving then.

2/4 $t = 1.5 s$ : $y = 31.25 m$ , $v = 0$ , but $a$ is still $- 10 m / s^{2}$ .

Fig. 2.8Take

y

upward from the water:

y_{0} = 20 m

v_{0} = + 15 m / s

a = - 10 m / s^{2}

.NCERT Fig. 2.6 · p. 19

A cliff 20 metres high stands at the left above a river, with a height scale in metres on the right. A dashed vertical path goes up from the cliff edge and back down to the water. The acceleration arrow, minus 10 metres per second squared, points down throughout. Step 1: at t equals 0 the stone leaves the edge with velocity plus 15 metres per second, arrow up. Step 2: at 1.5 seconds it is at the top, 31.25 metres above the river, with v equal to zero. Step 3: at 3.0 seconds it passes the cliff edge going down at minus 15 metres per second. Step 4: at 4.0 seconds it reaches the water at minus 25 metres per second.

Try it yourself first. Then reveal the solution one step at a time.

Example 2.4NCERT Example 2.4 · p. 19

Free fall from rest

Describe the motion of a body released from rest near the Earth's surface, ignoring air resistance. Find its velocity and position $0.5$ , $1$ , $2$ and $3 s$ after release.

Try it yourself first. Then reveal the solution one step at a time.

Drop a coin and a cricket ball together from a balcony: if air made no difference they would land together. Near the Earth every body falling under gravity alone gains speed at the same rate, $g \approx 9.8 m / s^{2}$ downward. Such motion is free fall (मुक्त पतन). $g$ weakens only slowly with height (by about 0.3 % for every 10 km), so for everyday falls it is constant and free fall is motion with constant acceleration.

Table 2.2Free fall from rest, up positive,

g = 9.8 m / s^{2}

$t$ (s)	$v$ (m/s)	$y$ (m)
0.5	−4.9	−1.23
1.0	−9.8	−4.9
2.0	−19.6	−19.6
3.0	−29.4	−44.1

Fig. 2.9 shows the three graphs of free fall: a flat $a$ – $t$ line, a straight $v$ – $t$ line with slope $- g$ , and a $y$ – $t$ curve that bends ever more steeply downward.

2/3Velocity: a straight line, $v = - 9.8 t$ .

Fig. 2.9Free fall, up positive: constant

a = - g

, linear

v

, parabolic

y

.NCERT Fig. 2.7 · p. 20

Graphs against time from 0 to 3 seconds, drawn below a time axis at the top because all values are negative. Step 1: acceleration is a flat line at minus 9.8 metres per second squared. Step 2: velocity is a straight line from 0 falling to minus 29.4 metres per second at 3 seconds. Step 3: position is a curve that starts flat and bends steeply downward, reaching minus 44.1 metres at 3 seconds.

Example 2.5NCERT Example 2.5 · p. 20

Galileo's rule of odd numbers

Galileo found that, starting from rest, the distances a falling body covers in successive equal stretches of time go as the odd numbers $1, 3, 5, 7, \dots$ Prove this rule.

Try it yourself first. Then reveal the solution one step at a time.

4/4Then 7 units (total 16 $= 4^{2}$ ).

Fig. 2.10Equal times, growing gaps:

1 : 3 : 5 : 7

, with running totals

1, 4, 9, 16

.NCERT Table 2.2 · p. 20

A vertical scale with a ball shown at the instants 0, tau, 2 tau, 3 tau and 4 tau after release. The ball positions are 0, 1, 4, 9 and 16 units below the start. Step 1 shows the first two positions and the first gap, 1 unit. Step 2 adds the third position and a gap of 3 units. Step 3 adds a gap of 5 units. Step 4 adds a gap of 7 units and the summary: distances in equal intervals are 1 : 3 : 5 : 7, totalling 16 units after four intervals.

Try this

Galileo's slope, with a metronome

You’ll need

a 2 m curtain rail or grooved plank
a marble or small steel ball
a book to raise one end
a phone metronome app
chalk or sticky notes, a measuring tape

Predict first

If the ball rolls

4 c m

during the first beat, how far should it roll during the third beat?

Galileo slowed free fall down with a slope. You can repeat his test with a ball and a steady beat.

Raise one end of a straight 2 m curtain rail or grooved plank by about 5 cm on a book, so that a ball released at the top takes about 5 s to reach the bottom.
Start a metronome app at about 60 beats per minute. Release the ball from rest exactly on a beat.
A partner marks the ball's position with chalk or a sticky note on each of the next four beats. Repeat a few times to settle each mark.
Measure each mark from the start, then the gap between neighbouring marks, and divide every gap by the first one.

Table 2.3

Beat	From start (cm)	Gap (cm)	Gap ÷ first
1
2
3
4

Your entries are saved automatically.

Example 2.6NCERT Example 2.6 · p. 20

Stopping distance

Measure from the instant the brakes bite to the instant the vehicle halts: that stretch of road is its stopping distance. (a) Find it in terms of the initial speed $v_{0}$ and a steady deceleration of size $∣ a ∣$ . (b) A loaded bus brakes at $6.0 m / s^{2}$ . Find its stopping distance at $15 m / s$ ( $54 k m / h$ ) and at twice that speed.

Try it yourself first. Then reveal the solution one step at a time.

On a real road the brakes do not act at once. During the driver's reaction time $t_{r}$ the vehicle keeps its full speed, adding $v_{0} t_{r}$ to the distance. Fig. 2.11 includes this for $t_{r} = 0.75 s$ : doubling the speed doubles the reaction distance and quadruples the braking distance.

2/2Add $v_{0} t_{r}$ for $t_{r} = 0.75 s$ : totals $30 m$ and $97.5 m$ .

Fig. 2.11Braking distance grows as

v_{0}^{2}

; reaction distance as

v_{0}

Two lanes of road, each with a car at the left. Upper lane: 15 metres per second; lower lane: 30 metres per second. Step 1 shows only the braking zones: 18.75 metres in the upper lane and 75 metres, four times as long, in the lower lane. Step 2 inserts a dashed reaction zone in front of each braking zone, 11.25 metres and 22.5 metres for a reaction time of 0.75 seconds, pushing the braking zones further along, and gives the totals 30 metres and 97.5 metres.

Example 2.7NCERT Example 2.7 · p. 21

Reaction time from a falling ruler

A friend holds a ruler vertically with its zero mark level with your open thumb and finger, then lets go without warning. You catch it after it has fallen $17.6 c m$ . Estimate your reaction time ( $g = 9.8 m / s^{2}$ ).

Try it yourself first. Then reveal the solution one step at a time.

3/3Caught after $d = 17.6 c m$ : $t_{r} = \sqrt{2 d / g} \approx 0.19 s$ .

Fig. 2.12The distance the ruler falls before you catch it gives your reaction time.NCERT Fig. 2.8 · p. 21

A vertical ruler marked in centimetres, with its zero mark at the bottom level with two fingers drawn as rounded bars on either side. Step 1: the friend is about to let go; the zero mark is at your fingers. Step 2: the ruler has started to fall, shown moved down with an arrow labelled falls freely. Step 3: the fingers close on the ruler; it has fallen 17.6 centimetres, so the reading at the fingers is d equals 17.6 centimetres, which gives a reaction time of about 0.19 seconds.

Try this

Measure your own reaction time

You’ll need

a 30 cm ruler
a friend
a calculator

Predict first

Before you start: if your reaction time were twice as long, the ruler would fall

twice as farfour times as farthe same distancehalf as far

Rest your forearm on a table with your hand over the edge, thumb and index finger about 2 cm apart.
Your friend holds the ruler at the top so that its 0 cm mark is level with the top of your thumb.
With no countdown, your friend lets go. Catch the ruler as quickly as you can and read the mark at the top of your thumb.
Repeat five times, work out each reaction time, then swap roles.

Table 2.4

Trial	$d$ (cm)	$t_{r}$ (s)
1
2
3
4
5

Your entries are saved automatically.

InteractiveReaction-time ruler

A ruler-drop game. A vertical 30 centimetre ruler hangs with its zero mark between two fingers. After you press Get ready, the ruler is released at a random moment one to three seconds later and falls with g equal to 9.8 metres per second squared. Pressing Catch, the space bar or tapping the stage stops it; the widget reads the distance fallen in centimetres and converts it into a reaction time using t equals the square root of 2d over g. Your last five attempts and their average are listed in a table. Pressing too early counts as a false start, and a ruler that falls past 30 centimetres is a miss.

Fig. 2.13Press Get ready, wait for the ruler to drop, then catch it. The distance it fell gives your reaction time through

t_{r} = \sqrt{2 d / g}

2.5 Relative velocity

Sitting in a train at a station, you may have felt your train start to move, only to see that it was the train on the next track pulling out. What you see depends on how you are moving. Describing one moving object from another calls for relative velocity (सापेक्ष वेग).

Let objects A and B move along the same line with constant velocities $v_{A}$ and $v_{B}$ , so that $x_{A} (t) = x_{A} (0) + v_{A} t$ and $x_{B} (t) = x_{B} (0) + v_{B} t$ . Define the position of B relative to A as the gap $x_{B A} = x_{B} - x_{A}$ , measured from A to B. Subtracting the two equations gives

x_{B A} (t) = x_{B A} (0) + v_{B A} t

(2.10)

where

Relative velocity of B with respect to A

v_{B A} = v_{B} - v_{A}

(2.11)

is the velocity of B relative to A: the velocity A would measure for B if A regarded itself as at rest. Likewise $v_{A B} = v_{A} - v_{B} = - v_{B A}$ : each sees the other moving at the same speed in the opposite direction.

Fig. 2.14 shows this on an $x$ – $t$ graph. If $v_{A} = v_{B}$ , the lines are parallel, $v_{B A} = 0$ and the gap never changes: two cars side by side at the same speed seem to each other to be standing still. If the velocities differ but share a sign, one line is steeper and the gap changes at the rate $∣ v_{B} - v_{A} ∣$ ; where the lines cross, one overtakes the other. If the velocities have opposite signs, the relative speed is the sum of the two speeds, and the objects close in on each other quickly.

2/3 $v_{B A} = + 10 m / s$ : B closes the $200 m$ gap in $20 s$ .

Fig. 2.14Parallel lines: constant gap. Crossing lines: the vehicles are level at the crossing.

Position from 0 to 1000 metres against time from 0 to 40 seconds. Step 1: A starts at 200 metres and B at 0, both at 20 metres per second; the lines are parallel and a double arrow shows the gap staying 200 metres. Step 2: B now moves at 30 metres per second; its steeper line meets A's at 20 seconds and 600 metres, where B catches A. Step 3: A starts at 0 moving at 20 metres per second and B starts at 700 metres moving at minus 15 metres per second; the rising and falling lines cross at 20 seconds and 400 metres, where they meet.

Example

Overtaking and meeting on a highway

A bus moves along a straight highway at $72 k m / h$ . (a) A car $200 m$ behind the bus travels the same way at $108 k m / h$ . When and where does it draw level with the bus? (b) Later, a truck coming the other way at $54 k m / h$ is $700 m$ ahead of the bus. When and where do they pass?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Two trains run on parallel tracks in the same direction at

90 k m / h

and

60 k m / h

. At what speed does a passenger in the slower train see the faster one move past?

km/h

Show answer

Answer 30 km/h

v_{B A} = 90 - 60 = 30 k m / h

, forward.

Recap

Motion in a Straight Line: key points and summary

Motion along a line is described by one signed coordinate $x$ , measured from a chosen origin in a chosen positive direction (recap).
Displacement $Δ x = x_{2} - x_{1}$ may be positive, negative or zero; path length counts every metre covered, so average speed $\geq ∣ \overset{ˉ}{v} ∣$ (Fig. 2.1).
Instantaneous velocity $v = d x / d t$ is the limit of $Δ x / Δ t$ and the slope of the tangent to the $x$ – $t$ graph; speed is $∣ v ∣$ (Section 2.2).
Average acceleration is $Δ v / Δ t$ ; instantaneous acceleration $a = d v / d t$ is the slope of the $v$ – $t$ graph (Section 2.3).
Uniform motion: straight $x$ – $t$ line, horizontal $v$ – $t$ line. Constant acceleration: parabolic $x$ – $t$ curve, straight sloping $v$ – $t$ line (Fig. 2.3, Fig. 2.4).
An object speeds up when $v$ and $a$ have the same sign and slows down when they differ; $v = 0$ does not mean $a = 0$ (bead example).
The area under a $v$ – $t$ graph between two instants is the displacement; area below the time axis counts as negative (Fig. 2.5).
Constant acceleration: $v = v_{0} + a t$ , $x - x_{0} = v_{0} t + \frac{1}{2} a t^{2}$ , $v^{2} = v_{0}^{2} + 2 a (x - x_{0})$ , $x - x_{0} = \frac{1}{2} (v_{0} + v) t$ ((2.9a)).
Free fall, up positive: $a = - g$ ; from rest, distances in successive equal intervals are in the ratio $1 : 3 : 5 : \dots$ (Example 2.5).
Braking distance $v_{0}^{2} / 2 ∣ a ∣$ grows as the square of the speed; a falling ruler gives the reaction time as $\sqrt{2 d / g}$ (Example 2.6, Example 2.7).
The velocity of B relative to A is $v_{B A} = v_{B} - v_{A}$ , the rate at which the gap $x_{B} - x_{A}$ changes (Section 2.5).

Table 2.5Where each quantity lives on a graphNCERT Summary table · p. 22

Quantity	Signed?	On a graph	SI unit
Displacement $Δ x$	yes	signed area under $v$ – $t$	m
Path length	no	total area, all counted $+$	m
Average velocity $\overset{ˉ}{v}$	yes	slope of a chord on $x$ – $t$	$m s^{- 1}$
Velocity $v$	yes	slope of the tangent on $x$ – $t$	$m s^{- 1}$
Speed $∣ v ∣$	no	steepness of $x$ – $t$	$m s^{- 1}$
Average acceleration $\overset{ˉ}{a}$	yes	slope of a chord on $v$ – $t$	$m s^{- 2}$
Acceleration $a$	yes	slope of the tangent on $v$ – $t$	$m s^{- 2}$

Practice · 27 questions

Practice questions with answers (NCERT “Exercises”)

Questions 1–18 practise the skills of NCERT Exercises 2.1–2.18 in new situations; 19–27 are entrance-style practice (JEE Main and NEET pattern). Take $g = 9.8 m / s^{2}$ unless a question says otherwise.

1
Quick answerEasyNCERT Exercise 2.1
In which of these questions is it safe to shrink the moving body to a single point? Select all that apply.
How long a passenger train takes for its $350 k m$ run from Chennai to BengaluruHow fast the tip of a ceiling-fan blade movesHow long a skydiver takes to fall $1 k m$ How a spinning carrom striker rebounds off the edge of the board
Select all that apply
Show answer
Answer
(A) How long a passenger train takes for its $350 k m$ run from Chennai to Bengaluru
(C) How long a skydiver takes to fall $1 k m$
The train (a few hundred metres long, travelling hundreds of kilometres) and the skydiver (under 2 m tall, falling 1000 m) are tiny compared with the distances covered. For the fan blade and the spinning striker the question is about how different parts move, so size and shape matter.
2
Short answerMediumNCERT Exercise 2.2
Kabir and Riya cycle home from the same market M along one straight road. Their $x$ – $t$ graphs are shown (Kabir solid, Riya dashed); each graph turns flat once the rider is home.
Fig. 2.15
Position in metres from the market against time in minutes. Kabir's solid line starts at 0 at time 0 and rises steadily to 3000 metres at 20 minutes, then stays flat. Riya's dashed line starts at 0 at 4 minutes, rises more steeply to 2500 metres at 14 minutes, then stays flat. The lines cross at 10 minutes and 1500 metres, and Kabir's line crosses Riya's flat line at about 16.7 minutes.
1. Who lives farther from the market?
  RiyaKabirThey live equally far
  Show answer
  Answer (B) Kabir
  Kabir's graph levels off at $3000 m$ , Riya's at $2500 m$ .
2. Who leaves the market first, and by how long?
  Riya, by 4 minThey leave togetherKabir, by 4 min
  Show answer
  Answer (C) Kabir, by 4 min
  Kabir's line starts rising at $t = 0$ , Riya's at $t = 4 m i n$ .
3. Who rides faster? Find both speeds.
  Show solution
  Slopes: Kabir $3000 / 20 = 150 m / m i n$ ( $2.5 m / s$ ); Riya $2500 / 10 = 250 m / m i n$ (about $4.2 m / s$ ). Riya is faster.
4. The graphs cross twice. How many times does one rider overtake the other on the road?
  twiceoncenever
  Show solution
  Once. At $t = 10 m i n$ both are $1500 m$ from the market and the faster Riya passes Kabir. The second crossing, at about $16.7 m i n$ , is where Kabir's line meets Riya's flat line: Riya is already home and Kabir is simply riding past her house.
DiagramEasyNCERT Exercise 2.3
A milk vendor leaves home at 6:00 am and cycles at $12 k m / h$ along a straight road to a dairy $3.0 k m$ away. He waits there for 30 min, then rides home at $18 k m / h$ . Choose suitable scales and draw the $x$ – $t$ graph of his trip.
Show solution
AnswerCorners at (6:00, 0), (6:15, $3.0 k m$ ), (6:45, $3.0 k m$ ) and (6:55, 0).
Outward: $3.0 / 12 = 0.25 h = 15 m i n$ , so he reaches the dairy at 6:15 and leaves at 6:45. Return: $3.0 / 18 = \frac{1}{6} h = 10 m i n$ , home at 6:55.
Table 2.6
Time $x$ (km)
6:00 0
6:15 3.0
6:45 3.0
6:55 0
Scale: 1 cm for 5 min and 1 cm for 0.5 km. Join the points with straight lines: rising with slope $12 k m / h$ , flat while he waits, then falling more steeply with slope $- 18 k m / h$ .
NumericalMediumNCERT Exercise 2.4
A toy robot on a straight track repeats a pattern: 6 steps forward, then 4 steps back. Each step is $0.5 m$ long and takes $1 s$ . The track ends at a gap $6.0 m$ ahead of the start. Sketch the $x$ – $t$ graph and find when the robot reaches the gap.
s
Show solution
Each cycle lasts $10 s$ : $3.0 m$ forward and $2.0 m$ back, a net $1.0 m$ . The $x$ – $t$ graph is a zigzag: rising lines of slope $+ 0.5 m / s$ for 6 s and falling lines of slope $- 0.5 m / s$ for 4 s, each peak $1.0 m$ higher than the last.
After $n$ full cycles the robot is at $n$ metres, and during the next forward run it reaches $(n + 3)$ metres. The gap at $6.0 m$ is first reached when $n + 3 = 6$ : after 3 cycles (30 s) plus 6 forward steps.
$t = 30 + 6 = 36 s$
On the graph, the peaks ( $3$ , $4$ , $5$ , $6 m$ at $6$ , $16$ , $26$ , $36 s$ ) first touch $6.0 m$ at $36 s$ . Dividing $6.0 m$ by the net $0.1 m / s$ would wrongly give $60 s$ .
5
NumericalEasyNCERT Exercise 2.5
An aircraft touches down at $288 k m / h$ and rolls $1.6 k m$ along the runway before stopping. Assuming uniform deceleration, find:
1. the size of the deceleration
  m/s²
  Show solution
  $v_{0} = 288 / 3.6 = 80 m / s$ . From $0 = v_{0}^{2} - 2 ∣ a ∣ x$ :
  $∣ a ∣ = \frac{v_{0}^{2}}{2 x} = \frac{6400}{3200}$
  $∣ a ∣ = 2.0 m / s^{2}$
2. the time it takes to stop
  s
  Show solution
  $t = v_{0} / ∣ a ∣ = 80 / 2.0 = 40 s$ . Check: $\frac{1}{2} (80 + 0) (40) = 1600 m$ .
6
Short answerMediumNCERT Exercise 2.6
A stone is thrown vertically upward at $24.5 m / s$ and caught again at the same height. Ignore air resistance; take $g = 9.8 m / s^{2}$ .
1. The stone pauses for an instant at the top. Which pair describes it there?
  $v = 0$ , $a = 0$ $v = 24.5 m / s$ up, $a = 0$ $v = 0$ , $a = 9.8 m / s^{2}$ up $v = 0$ , $a = 9.8 m / s^{2}$ down
  Show answer
  Answer (D) $v = 0$ , $a = 9.8 m / s^{2}$ down
  The stone stops for an instant, but gravity still acts: $a = g$ , downward, all the time.
2. Put the origin at the top of the flight and start the clock there ( $t = 0$ ), with upward as positive. What signs do $x$ , $v$ and $a$ have before the top ( $t < 0$ ) and after it ( $t > 0$ )?
  Show solution
  Table 2.7
  $x$ $v$ $a$
  Before the top − + −
  After the top − − −
  Every other point of the flight is below the top, so $x < 0$ on both legs. On the way up the stone moves in the positive direction ( $v > 0$ ); on the way down, $v < 0$ . Gravity points down throughout: $a = - 9.8 m / s^{2}$ . (With down positive every sign would flip.)
3. How high does it rise, and how long after the throw is it caught?
  Show solution
  $h = \frac{v_{0}^{2}}{2 g} = \frac{{24.5}^{2}}{19.6}$
  $h \approx 30.6 m$
  Time to the top: $v_{0} / g = 2.5 s$ . The descent takes as long, so it is caught $5.0 s$ after the throw.
7
Short answerMediumNCERT Exercise 2.7
Is each statement about a particle moving along a straight line true or false? Give a reason or an example.
1. A particle that is momentarily at rest must have zero acceleration at that instant.
  TrueFalse
  Show answer
  Answer False
  False. A ball at the top of its flight has $v = 0$ but $a = g$ .
2. If a particle's speed is zero at some instant, its velocity is zero at that instant.
  TrueFalse
  Show answer
  Answer True
  True. Speed is $∣ v ∣$ , and $∣ v ∣ = 0$ only when $v = 0$ .
3. If a particle covers the same path length in every second of an interval, its velocity stays constant over that interval.
  TrueFalse
  Show answer
  Answer False
  False. A puck sliding on smooth ice between two boards covers equal distances every second, but each elastic bounce reverses its velocity, which takes a large acceleration for a brief moment.
4. A particle whose velocity and acceleration are both negative must be slowing down.
  TrueFalse
  Show answer
  Answer False
  False. Same signs mean speeding up: with up positive, a falling stone has $v < 0$ and $a = - 9.8 m / s^{2}$ , and it gets faster. Slowing down needs opposite signs.
8
DiagramHardNCERT Exercise 2.8
A ball is dropped from $20 m$ onto a hard floor. At every bounce it leaves the floor with 80 % of the speed it arrived with. Take $g = 10 m / s^{2}$ and treat each bounce as instantaneous. Sketch its speed–time graph from $t = 0$ to $t = 8 s$ .
1. When does the ball first hit the floor, and how fast is it moving?
  Show solution
  From $20 = \frac{1}{2} (10) t^{2}$ , $t = 2.0 s$ ; then $v = g t = 20 m / s$ .
2. When does it hit the floor the second time?
  s
  Show solution
  It rebounds at $0.8 \times 20 = 16 m / s$ , rises for $1.6 s$ and falls for $1.6 s$ : second impact at $2.0 + 3.2 = 5.2 s$ , again at $16 m / s$ .
3. Describe the speed–time graph up to $t = 8 s$ .
  Show solution
  Rebound speeds: $16$ , $12.8$ , $10.24 m / s$ . Flights last $3.2 s$ and then $2.56 s$ , so impacts come at $2.0$ , $5.2$ and $7.76 s$ .
  Table 2.8
  t (s) speed (m/s) event
  0 0 released
  2.0 20 → 16 bounce 1
  3.6 0 top, 12.8 m
  5.2 16 → 12.8 bounce 2
  6.48 0 top, 8.19 m
  7.76 12.8 → 10.24 bounce 3
  8.0 7.84 rising
  Between these points the speed changes at $10 m / s$ every second, falling to zero on each rise and growing on each fall. The graph is a row of shrinking 'V' shapes with a sudden drop at every bounce.
9
Long answerMediumNCERT Exercise 2.9
A lift starts at the ground floor, rises to the 8th floor ( $32 m$ up) and then comes down to the 3rd floor ( $12 m$ up), taking $40 s$ in all. Use it to explain the difference between:
1. the size of the displacement and the path length for the trip;
  Show solution
  With up positive, the displacement is $+ 12 m$ , but the lift travels $32 + 20 = 52 m$ . In general the path length is the sum of the sizes of the small displacements, $\sum ∣ Δ x_{i} ∣$ , while the displacement is their signed sum, $\sum Δ x_{i}$ . The size of a sum never exceeds the sum of the sizes, so path length $\geq ∣ Δ x ∣$ . They are equal only if every $Δ x_{i}$ has the same sign: the motion never reverses.
2. the size of the average velocity and the average speed for the whole trip.
  Show solution
  Divide both by the same $Δ t = 40 s$ : $∣ \overset{ˉ}{v} ∣ = 12 / 40 = 0.30 m / s$ , while the average speed is $52 / 40 = 1.3 m / s$ . Since path length $\geq ∣ Δ x ∣$ , average speed $\geq ∣ \overset{ˉ}{v} ∣$ , with equality only for motion that does not turn back.
10
NumericalMediumNCERT Exercise 2.10
Meera jogs from home along a straight road to a park gate $1.8 k m$ away at $9.0 k m / h$ . Finding the gate shut, she turns at once and walks home at $6.0 k m / h$ . Find the magnitude of her average velocity and her average speed over these intervals:
1. 0 to 12 min
  Show solution
  She reaches the gate at $1.8 / 9.0 = 0.20 h = 12 m i n$ , with no reversal yet: $∣ \overset{ˉ}{v} ∣$ and the average speed are both $9.0 k m / h$ .
2. 0 to 30 min
  Show solution
  The walk back takes $1.8 / 6.0 = 0.30 h = 18 m i n$ , so she is home at 30 min. Displacement is zero, so $∣ \overset{ˉ}{v} ∣ = 0$ ; the average speed is $3.6 k m / 0.50 h = 7.2 k m / h$ . Calling her average speed zero after a 3.6 km outing would be absurd, which is why average speed uses path length.
3. 0 to 20 min (enter the average speed)
  km/h
  Show solution
  After the gate she walks for 8 min, covering $6.0 \times 8 / 60 = 0.80 k m$ , so she is $1.0 k m$ from home. In $20 m i n = \frac{1}{3} h$ : $∣ \overset{ˉ}{v} ∣ = 3.0 k m / h$ ; her path is $2.6 k m$ , so the average speed is $7.8 k m / h$ .
Short answerMediumNCERT Exercise 2.11
Over a finite interval the average speed can exceed the size of the average velocity. Explain why, at any single instant, the speed and the size of the velocity are always equal.
Show solution
The average speed exceeds $∣ \overset{ˉ}{v} ∣$ only because the object may reverse during the interval, adding path length but not displacement. As $Δ t \to 0$ there is no time for a reversal, so the path length in the interval becomes equal to $∣ Δ x ∣$ (at a turning point both shrink faster than $Δ t$ ). Dividing by $Δ t$ and taking the limit, the instantaneous speed equals $∣ d x / d t ∣ = ∣ v ∣$ .
12
Quick answerMediumNCERT Exercise 2.12
Which of these graphs show something that no particle moving along a straight line could do? Select all that apply.
Fig. 2.16
Four small graphs. (a) Position against time: a curve shaped like a sideways U opening to the right, so some times have two positions. (b) Speed against time: a wave that dips below the time axis. (c) Path length against time: rises and then stays level. (d) Position against time: a sloping line with a vertical jump in the middle, then another sloping line.
(a)(b)(c)(d)
Select all that apply
Show answer
Answer
(A) (a)
(B) (b)
(D) (d)
(a) gives two positions at the same instant. (b) shows a negative speed, but speed is a magnitude. (d) jumps from one position to another in zero time, an infinite velocity. (c) is possible: path length grows, then stays fixed while the particle is at rest.
Short answerMediumNCERT Exercise 2.13
The figure shows the $x$ – $t$ graph of a particle: a straight line for $t < 0$ and the curve $x = 4 t - t^{2}$ (SI units) for $t \geq 0$ . A student says the particle moves along a straight path before $t = 0$ and along a curved path after it. Is that right? Suggest a real situation the graph could describe.
Fig. 2.17
Position in metres against time in seconds from minus 2 to 4. For negative times the graph is a straight line rising with slope 4 metres per second and reaching the origin at t equals 0. After that it follows a curve that rises more and more slowly to a peak of 4 metres at 2 seconds, then falls back to 0 at 4 seconds.
Show solution
No. The particle moves along the $x$ -axis all the time; the graph only shows how $x$ changes with $t$ . Before $t = 0$ the slope is constant, $v = 4 m / s$ . After it, $v = 4 - 2 t$ and $a = - 2 m / s^{2}$ : the particle slows, stops at $t = 2 s$ at $x = 4 m$ and comes back, passing $x = 0$ at $t = 4 s$ . A remote-controlled car cruising at $4 m / s$ , switched at $t = 0$ to a steady backward push of $2 m / s^{2}$ , would behave like this: it brakes, stops and then reverses.
NumericalMediumNCERT Exercise 2.14
During a film shoot, a camera jeep moving at $36 k m / h$ fires a paintball forward at a stunt car racing ahead on the same road at $108 k m / h$ . The paintball leaves the marker at $80 m / s$ relative to the jeep. Ignoring air resistance, with what speed does it hit the car? (Give the paintball's speed as seen from the car, since that decides how hard it hits.)
m/s
Show solution
Forward is positive. Jeep $36 k m / h = 10 m / s$ ; car $108 k m / h = 30 m / s$ . The paintball relative to the ground: $10 + 80 = 90 m / s$ . Relative to the car: $90 - 30 = 60 m / s$ .
15
Short answerMediumNCERT Exercise 2.15
Describe a real motion that could produce each graph (a)–(c).
Fig. 2.18
Three small graphs side by side. (a) Velocity against time: zero, then a sudden vertical jump to a large value, then a straight-line fall back to zero, after which it stays zero. (b) Position against time: a curve that rises steeply at first, then more and more gently, and finally levels off. (c) Acceleration against time: a constant negative value except for one short, tall positive spike.
1. Graph (a), velocity against time.
  Show solution
  A hockey puck at rest is struck by a stick (an almost instant jump in velocity), then slides over rough ground, slowing at a steady rate until it stops and stays at rest.
2. Graph (b), position against time.
  Show solution
  A car braking to a stop, or a ball rolling to rest on a carpet: the position rises quickly at first, then more slowly, and finally stays fixed as the slope (velocity) falls to zero.
3. Graph (c), acceleration against time.
  Show solution
  A ball falling freely (up positive, $a = - g$ ) that bounces once off a hard floor: during the brief contact the floor pushes it strongly upward, giving the large positive spike, after which $a = - g$ again.
16
Short answerMediumNCERT Exercise 2.16
The figure shows the $x$ – $t$ graph of a particle oscillating along a line (simple harmonic motion, which you will study later in Class 11): $x = A \cos (π t / 2)$ with $t$ in seconds. Give the signs of $x$ , $v$ and $a$ at each time.
Fig. 2.19
Position against time from minus 4 to 4 seconds: a smooth cosine wave of amplitude A. It is at its maximum A at t equals 0, crosses zero at plus and minus 1 second, reaches minus A at plus and minus 2 seconds, crosses zero again at plus and minus 3 seconds and returns to A at plus and minus 4 seconds.
1. $t = 0.5 s$
  $x > 0, v < 0, a < 0$ $x > 0, v > 0, a < 0$ $x < 0, v < 0, a > 0$ $x < 0, v > 0, a > 0$
  Show answer
  Answer (A) $x > 0, v < 0, a < 0$
  Above the axis ( $x > 0$ ) and falling ( $v < 0$ ); $a = - (π / 2)^{2} x < 0$ .
2. $t = 1.5 s$
  $x > 0, v < 0, a < 0$ $x > 0, v > 0, a < 0$ $x < 0, v < 0, a > 0$ $x < 0, v > 0, a > 0$
  Show answer
  Answer (C) $x < 0, v < 0, a > 0$
  Below the axis and still falling: $x < 0$ , $v < 0$ , so $a > 0$ .
3. $t = - 0.5 s$
  $x > 0, v < 0, a < 0$ $x > 0, v > 0, a < 0$ $x < 0, v < 0, a > 0$ $x < 0, v > 0, a > 0$
  Show answer
  Answer (B) $x > 0, v > 0, a < 0$
  Above the axis and rising towards the peak at $t = 0$ : $x > 0$ , $v > 0$ , $a < 0$ .
17
Short answerMediumNCERT Exercise 2.17
The figure shows the $x$ – $t$ graph of a particle, with three equal time intervals shaded. Answer from the graph.
Fig. 2.20
A smooth position-time curve. In shaded interval 1 the curve rises at a moderate slope. Later it reaches a peak and then, in shaded interval 2, falls very steeply through the time axis. After a trough below the axis it rises gently again, and in shaded interval 3 it climbs only slowly.
1. Which interval shows the highest average speed?
  interval 1interval 2interval 3
  Show answer
  Answer (B) interval 2
  Interval 2 has the steepest slope, so the largest |Δx| in the same time.
2. Which interval shows the lowest average speed?
  interval 1interval 2interval 3
  Show answer
  Answer (C) interval 3
  Interval 3 has the gentlest slope.
3. What are the signs of the average velocity in intervals 1, 2 and 3?
  $+, -, +$ $+, +, -$ $-, +, +$ $+, -, -$
  Show answer
  Answer (A) $+, -, +$
  $x$ increases in 1, decreases in 2 and increases in 3. The curve does not turn back inside any interval, so each average speed equals $∣ \overset{ˉ}{v} ∣$ .
18
Short answerHardNCERT Exercise 2.18
A particle never changes its direction of travel; call that direction positive. Its speed is graphed against time below, with three shaded strips of equal width and four marked instants P, Q, R and S.
Fig. 2.21
A smooth speed-time curve that stays above the time axis. It dips to a shallow minimum at P, rises steeply through shaded interval 1 to a rounded maximum at Q, falls gently just after Q in shaded interval 2, then falls more steeply through R in shaded interval 3, levels into a trough, and rises to a second, lower rounded maximum at S.
1. Which shaded interval has the largest average acceleration, ignoring its sign?
  interval 1interval 2interval 3
  Show answer
  Answer (A) interval 1
  The speed changes most, by a steep rise, across interval 1.
2. Which interval has the highest average speed?
  interval 1interval 2interval 3
  Show answer
  Answer (B) interval 2
  Interval 2 lies near the highest part of the curve.
3. Give the signs of $v$ and $a$ in each interval.
  Show solution
  $v > 0$ in all three, since the motion is always in the positive direction. $a > 0$ in interval 1 (speed rising) and $a < 0$ in intervals 2 and 3 (speed falling, gently in 2 and faster in 3).
4. What is the acceleration at P, Q, R and S?
  Show solution
  At P (a minimum), Q and S (maxima) the tangent is horizontal, so $a = 0$ . At R the curve is falling steeply, so $a$ is negative there and not zero.
Quick answerMedium
A particle moves along the $x$ -axis with $x = 3 t^{3} - 9 t + 2$ (SI units). The distance it travels in the first $2 s$ is
$6 m$ $12 m$ $18 m$ $24 m$
Show answer
Answer (C) $18 m$
$v = 9 t^{2} - 9 = 0$ at $t = 1 s$ , where the particle turns back. $x (0) = 2$ , $x (1) = - 4$ , $x (2) = 8 m$ , so the distance is $6 + 12 = 18 m$ . The displacement is only $6 m$ .
Quick answerEasy
Assertion (A): A body can have zero velocity and non-zero acceleration at the same instant. Reason (R): Acceleration depends on how fast the velocity is changing, not on the value of the velocity.
Both A and R are true, and R correctly explains ABoth A and R are true, but R does not explain AA is true but R is falseA is false but R is true
Show answer
Answer (A) Both A and R are true, and R correctly explains A
At the top of a vertical throw $v = 0$ while $d v / d t = - g$ . R gives exactly the reason.
Quick answerMedium
Assertion (A): At the highest point of its flight, a ball thrown vertically upward has an acceleration $g$ directed downward. Reason (R): At the highest point the ball is momentarily at rest, so no force acts on it there.
Both A and R are true, and R correctly explains ABoth A and R are true, but R does not explain AA is true but R is falseA is false but R is true
Show answer
Answer (C) A is true but R is false
A is true. R is false: gravity acts at every instant, including the top. Being at rest for an instant does not remove the force.
NumericalMedium
A stone dropped from the top of a tower covers $25 m$ in the last second of its fall. Find the height of the tower ( $g = 10 m / s^{2}$ ).
m
Show answer
Answer 45 m
$5 (2 n - 1) = 25$ gives $n = 3$ , so the fall lasts $3 s$ and $h = \frac{1}{2} (10) (3)^{2} = 45 m$ .
NumericalEasy
A car accelerates uniformly from rest to $20 m / s$ in $10 s$ , runs at that speed for $20 s$ , then brakes uniformly to rest in $5 s$ . Using the area under its $v$ – $t$ graph, find the total distance covered.
m
Show answer
Answer 550 m
$\frac{1}{2} (10) (20) + 20 (20) + \frac{1}{2} (5) (20) = 100 + 400 + 50 = 550 m$ .
Quick answerMedium
Two stones are dropped from the same height, the second one $1.0 s$ after the first. While both are falling, the distance between them
stays constantincreases at a steady rateincreases ever faster, as $t^{2}$ decreases
Show answer
Answer (B) increases at a steady rate
Both accelerate at $g$ , so the relative acceleration is zero and the relative velocity stays $g \times 1.0 s = 9.8 m / s$ . The gap therefore grows linearly with time.
25
Quick answerEasy
Match each graph with the motion it shows.
Tap an item on the left, then its match on the right.
Show answer
Answer
$x$ – $t$ : horizontal line : at rest
$v$ – $t$ : rising line through the origin : from rest, constant positive acceleration
$a$ – $t$ : horizontal line at $- g$ : free fall, up positive
$x$ – $t$ : straight line, negative slope : constant velocity, negative direction
Read the quantity on the vertical axis first: a flat x–t line means x is not changing; a straight v–t line through the origin means v grows steadily from zero.
NumericalHard
A particle's acceleration is $a = 6 t$ (SI units) and its velocity at $t = 0$ is $2 m / s$ . Find its velocity at $t = 2 s$ .
m/s
Show answer
Answer 14 m/s
$v = 2 + \int_{0}^{t} 6 t^{'} d t^{'} = 2 + 3 t^{2}$ , so $v (2) = 14 m / s$ . Using $v = v_{0} + a t$ with $a (2) = 12$ would wrongly give $26 m / s$ .
Quick answerEasy
A car moving at $36 k m / h$ can be stopped in $8.0 m$ . With the same braking, its stopping distance at $72 k m / h$ is
$64 m$ $32 m$ $24 m$ $16 m$
Show answer
Answer (B) $32 m$
$d_{s} \propto v_{0}^{2}$ : doubling the speed multiplies $8.0 m$ by 4.

Time	$x$ (km)
6:00	0
6:15	3.0
6:45	3.0
6:55	0

	$x$	$v$	$a$
Before the top	−	+	−
After the top	−	−	−

t (s)	speed (m/s)	event
0	0	released
2.0	20 → 16	bounce 1
3.6	0	top, 12.8 m
5.2	16 → 12.8	bounce 2
6.48	0	top, 8.19 m
7.76	12.8 → 10.24	bounce 3
8.0	7.84	rising

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Quick check: MCQs on Motion in a Straight Line

Position, displacement and average velocity: a recap3 questions
2.2 Instantaneous velocity and speed3 questions
2.3 Acceleration4 questions
2.4 Kinematic equations for uniformly accelerated motion4 questions
2.5 Relative velocity3 questions

Glossary · 15

Key terms (English–Hindi)

accelerationत्वरणtvaran: Rate of change of velocity with time: average $Δ v / Δ t$ , instantaneous $d v / d t$ ; SI unit $m s^{- 2}$ .First seen in the chapter →
average speedऔसत चालausat chaal: Total path length divided by the time taken; never less than the size of the average velocity.First seen in the chapter →
average velocityऔसत वेगausat veg: Displacement divided by the time taken, $\overset{ˉ}{v} = Δ x / Δ t$ ; the slope of a chord on the x–t graph.First seen in the chapter →

displacementविस्थापनvisthapan: The change in position, $Δ x = x_{2} - x_{1}$ ; its sign gives the direction along the chosen axis.First seen in the chapter →

free fallमुक्त पतनmukt patan: Motion under gravity alone, air resistance neglected; near the Earth the acceleration is $g \approx 9.8 m / s^{2}$ downward.First seen in the chapter →

instantaneous velocityतात्क्षणिक वेगtatkshanik veg: The value the average velocity approaches as the time interval shrinks to zero, $v = d x / d t$ ; graphically, the tangent slope on the x–t graph.First seen in the chapter →

kinematicsशुद्धगतिकीshuddhgatiki: The description of motion (position, velocity, acceleration) without asking what causes it.First seen in the chapter →

path lengthपथ-लंबाईpath-lambai: The total distance actually travelled along the route; never negative and never less than the size of the displacement.First seen in the chapter →
point objectबिंदु वस्तुbindu vastu: A body whose size is negligible compared with the distance it moves in the time studied, so it can be treated as a single point.First seen in the chapter →

reaction time: The delay between noticing a signal and beginning to respond to it.First seen in the chapter →
rectilinear motionसरल रेखीय गतिsaral rekhiya gati: Motion along one straight line, so a single coordinate $x$ gives the position.First seen in the chapter →
relative velocityसापेक्ष वेगsapeksh veg: The velocity of one object as measured from another; along a line, $v_{B A} = v_{B} - v_{A}$ .First seen in the chapter →

speedचालchaal: The magnitude of the instantaneous velocity, $∣ v ∣$ .First seen in the chapter →
stopping distance: The distance a vehicle covers from the moment the brakes act until it stops; $v_{0}^{2} / 2 ∣ a ∣$ for steady braking.First seen in the chapter →

uniformly accelerated motionएकसमान त्वरित गतिeksaman tvarit gati: Motion whose acceleration keeps the same size and direction, so its v–t graph is straight.First seen in the chapter →

2.1 Introduction

Position, displacement and average velocity: a recap

2.2 Instantaneous velocity and speed

2.3 Acceleration

2.4 Kinematic equations for uniformly accelerated motion

2.5 Relative velocity

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