PhysicsCBSE Class 11 · Chapter 2

Motion in a Straight Line Class 11: Notes, Graphs and Solved Numericals

Updated 14 October 2026 · Fact-checked

Motion in a Straight Line is Chapter 2 of CBSE Class 11 Physics. It covers frame of reference, average and instantaneous velocity and speed, acceleration, position-time and velocity-time graphs, elementary differentiation and integration for motion, and the equations of uniformly accelerated motion derived by graphs and by calculus.

What Motion in a Straight Line is about

Kinematics describes motion without asking what causes it. This chapter keeps things in one dimension: a car on a straight road, a ball thrown straight up, a stone dropped down a well. You first fix a frame of reference (an origin, a direction taken as positive and a clock), so that position x is just a number with a sign.

From position you build velocity, the rate of change of position, and from velocity you build acceleration, the rate of change of velocity. The chapter introduces the calculus you need: velocity is the derivative $v = \frac{d x}{d t}$ , acceleration is $a = \frac{d v}{d t}$ , and going backwards, displacement is the integral of velocity (the area under the v-t graph).

For constant acceleration you get the three kinematic equations, $v = u + a t$ , $s = u t + \frac{1}{2} a t^{2}$ and $v^{2} = u^{2} + 2 a s$ , which you must be able to derive both from a v-t graph and by calculus. They describe free fall, stopping distances of vehicles and reaction time. Relative velocity has been removed from the 2026-27 syllabus. Everything here is used again in Motion in a Plane (projectiles) and in every mechanics chapter after it, and it is basic JEE and NEET material.

Weightage: This chapter is part of Unit II, Kinematics. The syllabus groups Unit II with Unit I, Physical World and Measurement and Unit III, Laws of Motion: together they carry 23 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit II Kinematics

Key concepts

Frame of reference, position, displacement and distance

NCERT section 2.1

To describe motion we need a frame of reference: a reference point (origin), axes with a chosen positive direction, and a clock. For motion along a line we use only the x-axis; positions to the right of O are positive and to the left negative.

Displacement is the change in position, $Δ x = x_{2} - x_{1}$ . It has a sign (direction) and can be zero even when the object has moved.
Path length (distance) is the total length actually travelled. It is never negative and is never less than the size of the displacement.

An object is treated as a point object when the distance it moves is much larger than its own size (a car on a 100 km trip).

Motion is uniform if equal displacements happen in equal intervals of time, however small; otherwise it is non-uniform.

Worked example

A girl walks 60 m east along a straight road and then 20 m west, taking 40 s in all. Taking east as positive, find (a) her displacement, (b) the distance she covers. [1 mark]

Try it first, then show the solutionSolution

(a) Final position $= + 60 - 20 = + 40$ m, so displacement $= + 40$ m (40 m east)
(b) Distance $= 60 + 20 = 80$ m

Answer: (a) 40 m east (b) 80 m

Average velocity, average speed and the x-t graph

NCERT section 2.1 and 2.2

$Average velocity = \frac{displacement}{time} = \frac{x _{2} - x _{1}}{t _{2} - t _{1}}$ and $average speed = \frac{total path length}{total time}$ .

Average speed is always at least as large as the magnitude of the average velocity; they are equal only if the object never turns back.

On a position-time (x-t) graph, the average velocity over an interval is the slope of the chord joining the two points.

Object at rest: horizontal line.
Uniform motion: straight sloping line (slope = velocity).
Accelerated motion: a curve; an upward-curving graph means increasing velocity.

Useful result: for two equal distances covered at speeds $v_{1}$ and $v_{2}$ , the average speed is $\frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$ (not the simple average).

Worked example

A car goes 30 km along a straight road at 60 km h $^{- 1}$ and comes back at 40 km h $^{- 1}$ . Find its average speed and average velocity for the round trip. [2 marks]

Try it first, then show the solutionSolution

Time out $= \frac{30}{60} = 0.5$ h; time back $= \frac{30}{40} = 0.75$ h; total 1.25 h
Average speed $= \frac{60 k m}{1.25 h} = 48 k m h^{- 1}$ (check: $\frac{2 \times 60 \times 40}{100} = 48$ )
It returns to the start, so displacement = 0 and average velocity = 0

Answer: Average speed 48 km h $^{- 1}$ ; average velocity zero

Instantaneous velocity as a derivative

NCERT section 2.2

Average velocity over a shrinking interval approaches a limit: the instantaneous velocity
$v = lim_{Δ t \to 0} \frac{Δ x}{Δ t} = \frac{d x}{d t}$

Geometrically it is the slope of the tangent to the x-t graph at that instant. Instantaneous speed is the magnitude of instantaneous velocity (unlike averages, the two always match in size at an instant).

Differentiation rules you need: $\frac{d}{d t} (t^{n}) = n t^{n - 1}$ , the derivative of a constant is 0, and the derivative of a sum is the sum of derivatives. So if $x = a + b t + c t^{2}$ , then $v = b + 2 c t$ .

Worked example

The position of a particle is $x = 4 t^{3} - 2 t + 1$ (x in m, t in s). Find its velocity at t = 2 s. [2 marks]

Try it first, then show the solutionSolution

$v = \frac{d x}{d t} = 12 t^{2} - 2$
At t = 2 s: $v = 12 \times 4 - 2 = 46 m s^{- 1}$

Answer: 46 m s $^{- 1}$

Acceleration and the v-t graph

NCERT section 2.3

Average acceleration $= \frac{v _{2} - v _{1}}{t _{2} - t _{1}}$ ; instantaneous acceleration $a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}}$ . SI unit: m s $^{- 2}$ .

The slope of a v-t graph is the acceleration.
The area under a v-t graph (taken with sign) is the displacement; this works for any motion, not only uniform acceleration.
Negative acceleration does not always mean slowing down. An object slows down when v and a have opposite signs and speeds up when they have the same sign.
Velocity can be zero while acceleration is not (a ball at the top of its flight), and acceleration can be zero while velocity is not (uniform motion).

A useful form when a depends on position: $a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x}$ .

Worked example

The velocity of a body is $v = 2 t + 3 t^{2}$ (m s $^{- 1}$ ). Find its acceleration at t = 1 s. [1 mark]

Try it first, then show the solutionSolution

$a = \frac{d v}{d t} = 2 + 6 t$
At t = 1 s: a = 8 m s $^{- 2}$

Answer: 8 m s $^{- 2}$

Kinematic equations: graphical derivation

NCERT section 2.4

For uniform acceleration the v-t graph is a straight line from (0, u) to (t, v).

1. Slope of the line $= a = \frac{v - u}{t}$ , so $v = u + a t$ .
2. Area under the line = rectangle + triangle $= u t + \frac{1}{2} (v - u) t = u t + \frac{1}{2} a t^{2}$ , so $s = u t + \frac{1}{2} a t^{2}$ .
3. The area is also a trapezium: $s = \frac{( u + v )}{2} t$ . Substituting $t = \frac{v - u}{a}$ gives $s = \frac{( v + u ) ( v - u )}{2 a}$ , so $v^{2} = u^{2} + 2 a s$ .

Here s is the displacement $x - x_{0}$ . Choose one direction as positive and keep every sign consistent: for a ball thrown up with up as positive, $a = - g$ .

The displacement in the nth second follows from $s_{n} - s_{n - 1}$ : $s_{n th} = u + \frac{a}{2} (2 n - 1)$ .

Worked example

A car moving at 72 km h $^{- 1}$ brakes with a uniform retardation of 4 m s $^{- 2}$ . Find the time and distance in which it stops. [2 marks]

Try it first, then show the solutionSolution

u = 72 km h $^{- 1}$ = 20 m s $^{- 1}$ , v = 0, a = −4 m s $^{- 2}$
$t = \frac{v - u}{a} = \frac{0 - 20}{- 4} = 5 s$
$s = \frac{v ^{2} - u ^{2}}{2 a} = \frac{0 - 400}{- 8} = 50 m$

Answer: 5 s and 50 m

Kinematic equations by calculus (integration)

NCERT section 2.4 (Example 2.2)

Integration undoes differentiation: $\int t^{n} d t = \frac{t ^{n + 1}}{n + 1}$ .

With a constant:

From $\frac{d v}{d t} = a$ : $\int_{u}^{v} d v = \int_{0}^{t} a d t$ gives $v - u = a t$ .
From $\frac{d x}{d t} = v = u + a t$ : $\int_{x_{0}}^{x} d x = \int_{0}^{t} (u + a t) d t$ gives $x - x_{0} = u t + \frac{1}{2} a t^{2}$ .
From $a = v \frac{d v}{d x}$ : $\int_{u}^{v} v d v = \int_{x_{0}}^{x} a d x$ gives $\frac{v ^{2} - u ^{2}}{2} = a (x - x_{0})$ , i.e. $v^{2} = u^{2} + 2 a (x - x_{0})$ .

The calculus method also works when acceleration changes with time: integrate a(t) to get v(t), then integrate v(t) to get x(t). Never use the three equations when a is not constant.

Worked example

A particle starts from rest at the origin with acceleration $a = 6 t$ (m s $^{- 2}$ ). Find its velocity and position at t = 2 s. [3 marks]

Try it first, then show the solutionSolution

$v = \int_{0}^{t} 6 t d t = 3 t^{2}$ (since v = 0 at t = 0)
$x = \int_{0}^{t} 3 t^{2} d t = t^{3}$ (since x = 0 at t = 0)
At t = 2 s: v = 12 m s $^{- 1}$ , x = 8 m

Answer: v = 12 m s $^{- 1}$ , x = 8 m

Free fall, vertical throws, stopping distance and reaction time

NCERT section 2.4 (Examples 2.3-2.7)

Near the Earth's surface, ignoring air resistance, every object has the same downward acceleration, g ≈ 9.8 m s $^{- 2}$ .

Free fall from rest (down positive): $v = g t$ , $y = \frac{1}{2} g t^{2}$ , $v^{2} = 2 g y$ .
Thrown up with speed u (up positive, a = −g): time to the top $= \frac{u}{g}$ , maximum height $= \frac{u ^{2}}{2 g}$ ; it returns to the same level after $\frac{2 u}{g}$ with the same speed u.
Galileo's law of odd numbers: from rest, distances in successive equal intervals are in the ratio 1 : 3 : 5 : 7 …
Stopping distance of a vehicle braking with retardation a: $d_{s} = \frac{u ^{2}}{2 a}$ , so doubling the speed makes it four times as long.
Reaction time $t_{r}$ : the vehicle keeps moving at speed u for $u t_{r}$ before the brakes act. A dropped-ruler test measures it: if the ruler falls d before you catch it, $t_{r} = \frac{2 d}{g}$ .

Worked example

A ball is thrown vertically up from the ground at 15 m s $^{- 1}$ . Find the maximum height and the total time of flight. (g = 10 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

$H = \frac{u ^{2}}{2 g} = \frac{225}{20} = 11.25 m$
Time to the top $= \frac{u}{g} = 1.5$ s, so total time $= 2 \times 1.5 = 3$ s

Answer: 11.25 m; 3 s

Physics in real life: Motion in a Straight Line

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Free fall

Suppose you drop a coin into a dry well and hear it hit the bottom 2.0 s later. Ignore the time taken by the sound and air resistance.

How deep is the well, and how fast is the coin moving when it hits the bottom? (g = 9.8 m s $^{- 2}$ )

Try it first, then show the solutionSolution

Depth $= \frac{1}{2} g t^{2} = \frac{1}{2} \times 9.8 \times 4 = 19.6 m$
Speed $= g t = 9.8 \times 2 = 19.6 m s^{- 1}$

Answer: 19.6 m deep; 19.6 m s $^{- 1}$

Memory hook: Half g t squared for how far, g t for how fast.

Average speed is not the average of speeds

Imagine you cycle to a coaching class 3 km away along a straight road, slowly because of a headwind, and ride back home along the same road faster with the wind behind you.

Suppose the speeds are 12 km h $^{- 1}$ going and 18 km h $^{- 1}$ returning. What is your average speed for the round trip?

Try it first, then show the solutionSolution

Equal distances, so $v_{a v} = \frac{2 v _{1} v _{2}}{v _{1} + v _{2}} = \frac{2 \times 12 \times 18}{30}$
$v_{a v} = 14.4 k m h^{- 1}$ (check: times 0.25 h + 0.1667 h = 0.4167 h for 6 km)

Answer: 14.4 km h $^{- 1}$ , not 15 km h $^{- 1}$

Memory hook: You spend longer at the slow speed, so it pulls the average down.

Reaction time

Suppose a friend holds a 30 cm ruler above your open fingers and lets go without warning; you catch it after it has fallen 20 cm.

Find your reaction time. If you were driving at 54 km h $^{- 1}$ , how far would the car travel in that time? (g = 9.8 m s $^{- 2}$ )

Try it first, then show the solutionSolution

$t_{r} = \frac{2 d}{g} = \frac{2 \times 0.20}{9.8} \approx 0.20 s$
54 km h $^{- 1}$ = 15 m s $^{- 1}$ , so distance $= 15 \times 0.20 = 3.0 m$

Answer: About 0.20 s; about 3 m before you even touch the brake

Memory hook: Reaction distance grows with speed, braking distance with speed squared.

Topics in this chapter

Motion in a Straight Line formula sheet

Average velocity and average speed: $\overset{v}{ˉ} = \frac{x _{2} - x _{1}}{t _{2} - t _{1}}, average speed = \frac{path length}{time}$
Instantaneous velocity and acceleration: $v = \frac{d x}{d t}, a = \frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}} = v \frac{d v}{d x}$
First equation of motion: $v = u + a t$
Second equation of motion: $s = u t + \frac{1}{2} a t^{2}$; s is the displacement $x - x_{0}$ .
Third equation of motion: $v^{2} = u^{2} + 2 a s$
Displacement from average of velocities: $s = \frac{( u + v )}{2} t$; Uniform acceleration only.
Displacement in the nth second: $s_{n th} = u + \frac{a}{2} (2 n - 1)$
Vertical throw (up positive, a = −g): $H = \frac{u ^{2}}{2 g}, t_{u p} = \frac{u}{g}, T = \frac{2 u}{g}$
Stopping distance and reaction time: $d_{s} = \frac{u ^{2}}{2 a}, t_{r} = \frac{2 d}{g}$
Average speed over two equal distances: $v_{a v} = \frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$

How the school exam asks Motion in a Straight Line

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Definitions and distinctions	Distance vs displacement, speed vs velocity, average vs instantaneous; can velocity be zero with non-zero acceleration?	Give a definition, a sign rule and a one-line example (ball at the top of its flight).
Calculus problems	x(t) given: find v and a at an instant, when the particle stops, distance in an interval; a(t) given: find v(t) and x(t).	Differentiate down (x → v → a), integrate up (a → v → x) with the initial values; find turning points from v = 0.
Graph reading	Interpret x-t, v-t and a-t graphs; find displacement from areas; sketch graphs for a described motion; spot impossible graphs.	x-t slope = v; v-t slope = a; v-t area = displacement; a body cannot be at two places at once.
Derivations	3-5 marks: derive the three equations of motion from a v-t graph or by calculus.	Draw the straight-line v-t graph, use slope and areas (rectangle + triangle, trapezium); or integrate dv = a dt and dx = v dt.
Free fall and vertical motion	Height, time, speed on hitting the ground; ball thrown up from a tower; last-second distance; two objects meeting.	Fix one positive direction; write each position with $y = y_{0} + u t - \frac{1}{2} g t^{2}$ ; solve the quadratic and reject negative time.
Stopping distance and reaction time	Total stopping distance with a reaction time; effect of doubling speed; ruler-drop test.	Total = $u t_{r} + \frac{u ^{2}}{2 a}$ ; braking distance ∝ $u^{2}$ .

Tricks to remember Motion in a Straight Line

Short memory aids. Say each one aloud once, then use it on a question the same day.

Slope goes down, area goes up
x-t slope gives v, v-t slope gives a (differentiate). v-t area gives displacement, a-t area gives change in velocity (integrate).
Use it when: Every graph question.
VUAT, SUAT, VUAS
The three equations each leave out one of v, s or t: $v = u + a t$ (no s), $s = u t + \frac{1}{2} a t^{2}$ (no v), $v^{2} = u^{2} + 2 a s$ (no t). Pick the one without the quantity you neither know nor want.
Use it when: Choosing an equation quickly.
Odd numbers fall
From rest: 1, 3, 5, 7 units in successive seconds; total after n seconds $\propto n^{2}$ .
Use it when: Galileo ratio and 'distance in nth second' MCQs.
Same signs speed up
v and a same sign: speeding up. Opposite signs: slowing down. The sign of a alone tells you nothing.
Use it when: Assertion-reason items on negative acceleration.
Up and down are mirror images
Without air resistance, time up = time down and speed at launch = speed on return to the same level.
Use it when: Vertical-throw problems.

Motion in a Straight Line: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Slope of an x-t graph · 1 mark · MCQExam level

The slope of the tangent to a position-time graph at an instant gives the:
(a) acceleration
(b) instantaneous velocity
(c) displacement
(d) average speed

Try it first, then show the solutionSolution

$v = d x / d t$ is the slope of the x-t curve at that instant.

Answer: (b) instantaneous velocity

Example 2: Zero velocity, non-zero acceleration · 1 mark · Assertion-reasonExam level

Assertion (A): A body can have zero velocity and still be accelerating.
Reason (R): At the highest point of a vertical throw, the velocity of a ball is zero while its acceleration is g.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

Both statements are true, and the ball at the top is exactly an example of A.

Answer: (a)

Example 3: When does the particle stop? · 2 marks · Very short answerHarder exam level

The position of a particle is $x = 2 t^{3} - 6 t^{2} + 4$ (x in m, t in s). Find when it is momentarily at rest and its acceleration at the later of those instants. [2 marks]

Try it first, then show the solutionSolution

$v = 6 t^{2} - 12 t = 6 t (t - 2) = 0$ at t = 0 and t = 2 s
$a = 12 t - 12$ ; at t = 2 s, $a = 12 m s^{- 2}$

Answer: At t = 0 and t = 2 s; a = 12 m s $^{- 2}$ at t = 2 s

Example 4: Last second of a fall · 3 marks · Short answerHarder exam level

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 $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

Let total time be T: distance in the last second $= \frac{1}{2} g T^{2} - \frac{1}{2} g (T - 1)^{2} = \frac{g}{2} (2 T - 1)$
$5 (2 T - 1) = 25 \Rightarrow T = 3 s$
$h = \frac{1}{2} \times 10 \times 9 = 45 m$

Answer: 45 m

Example 5: Distance from a v-t graph · 3 marks · Short answerHarder exam level

A train starts from rest, accelerates at 2 m s $^{- 2}$ for 5 s, runs at constant speed for 10 s and then decelerates uniformly to rest in 4 s. (a) Draw the v-t graph. (b) Find the total distance. (c) Find the deceleration. [3 marks]

Try it first, then show the solutionSolution

(a) Speed after 5 s = 2 × 5 = 10 m s $^{- 1}$ ; graph rises from 0 to 10 in 5 s, flat to 15 s, falls to 0 at 19 s
(b) Area $= \frac{1}{2} (5) (10) + 10 \times 10 + \frac{1}{2} (4) (10) = 25 + 100 + 20 = 145 m$
(c) Deceleration $= \frac{10}{4} = 2.5 m s^{- 2}$

Answer: 145 m; 2.5 m s $^{- 2}$

Example 6: Equations of motion and stopping distance · 5 marks · Long answerHarder exam level

(a) Derive $v^{2} = u^{2} + 2 a s$ from the v-t graph of uniformly accelerated motion. (b) A driver moving at 72 km h $^{- 1}$ sees a fallen tree 60 m ahead. Her reaction time is 0.5 s and the brakes give a retardation of 5 m s $^{- 2}$ . Does she stop in time? (c) What is the largest speed at which she could stop in 60 m? [5 marks]

Try it first, then show the solutionSolution

(a) Area of trapezium $s = \frac{( u + v )}{2} t$ and $t = \frac{v - u}{a}$ , so $s = \frac{( v + u ) ( v - u )}{2 a}$ , giving $v^{2} = u^{2} + 2 a s$
(b) u = 20 m s $^{- 1}$ . Reaction distance $= 20 \times 0.5 = 10 m$
Braking distance $= \frac{2 0 ^{2}}{2 \times 5} = 40 m$ ; total = 50 m < 60 m, so she stops 10 m short of the tree
(c) Need $0.5 u + \frac{u ^{2}}{10} = 60 \Rightarrow u^{2} + 5 u - 600 = 0$
$u = \frac{- 5 + 25 + 2400}{2} = \frac{- 5 + 49.24}{2} \approx 22.1 m s^{- 1}$ (about 80 km h $^{- 1}$ )

Answer: Yes, with 10 m to spare; up to about 22 m s $^{- 1}$ (80 km h $^{- 1}$ )

Example 7: Ball thrown up from a tower · 3 marks · Short answerHOTS / competency

A ball is thrown vertically upwards at 10 m s $^{- 1}$ from the top of a tower 40 m high. Find (a) the time it takes to reach the ground, (b) its speed on reaching the ground. (g = 10 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

Up positive, origin at the top: ground is at y = −40 m. $- 40 = 10 t - 5 t^{2} \Rightarrow t^{2} - 2 t - 8 = 0$
$(t - 4) (t + 2) = 0$ , so t = 4 s (reject −2 s)
$v = u - g t = 10 - 40 = - 30 m s^{- 1}$ ; check: $v^{2} = 100 + 2 \times 10 \times 40 = 900$

Answer: (a) 4 s (b) 30 m s $^{- 1}$ downwards

Example 8: Two balls meeting · 3 marks · Short answerHOTS / competency

A ball is dropped from a height of 80 m. At the same instant another ball is thrown straight up from the ground below it at 40 m s $^{- 1}$ . When and where do they meet? (g = 10 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

Heights above ground: $y_{1} = 80 - 5 t^{2}$ and $y_{2} = 40 t - 5 t^{2}$
They meet when $y_{1} = y_{2}$ : $80 = 40 t \Rightarrow t = 2 s$
Height $= 80 - 5 \times 4 = 60 m$ above the ground

Answer: After 2 s, 60 m above the ground

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Braking on a highway

Suppose a car is moving at 25 m s $^{- 1}$ on a straight road. The driver's reaction time is 0.6 s, and the brakes produce a steady retardation of 5 m s $^{- 2}$ .

(i) · 1 mark
How far does the car go during the reaction time?
Answer
Answer: 15 m
$25 \times 0.6 = 15 m$
(ii) · 1 mark
Find the braking distance.
Answer
Answer: 62.5 m
$\frac{2 5 ^{2}}{2 \times 5} = 62.5 m$
(iii) · 1 mark
Find the total stopping distance.
Answer
Answer: 77.5 m
15 + 62.5
(iv) · 1 mark
If the speed is doubled, what is the new braking distance?
Answer
Answer: 250 m
Braking distance ∝ $u^{2}$ : 4 × 62.5 m

A lift ride

Suppose a lift starts from rest, accelerates upwards at 1 m s $^{- 2}$ for 3 s, moves at constant speed for 10 s, and then slows uniformly to rest in 3 s.

(i) · 1 mark
What is the maximum speed of the lift?
Answer
Answer: 3 m s $^{- 1}$
$v = 0 + 1 \times 3$
(ii) · 1 mark
How far does it rise in the first 3 s?
Answer
Answer: 4.5 m
$\frac{1}{2} \times 1 \times 3^{2}$
(iii) · 1 mark
Find the total height risen.
Answer
Answer: 39 m
4.5 + 3 × 10 + 4.5
(iv) · 1 mark
Find the average speed for the whole ride.
Answer
Answer: About 2.4 m s $^{- 1}$
$\frac{39}{16} \approx 2.44$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

The position of a particle is $x = 5 t^{2} - 3 t$ (x in m, t in s). Find its velocity at t = 2 s and its acceleration. [2 marks] (2 marks)

Step	Marks
$v = \frac{d x}{d t} = 10 t - 3$ ; at t = 2 s, v = 17 m s $^{- 1}$	1
$a = \frac{d v}{d t} = 10 m s^{- 2}$ (constant)	1

Using calculus, derive $v^{2} = u^{2} + 2 a s$ for uniformly accelerated motion. [3 marks] (3 marks)

Step	Marks
$a = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t} = v \frac{d v}{d x}$ , so $v d v = a d x$	1
Integrate with limits: $\int_{u}^{v} v d v = a \int_{x_{0}}^{x} d x$ (a constant)	1
$\frac{v ^{2} - u ^{2}}{2} = a (x - x_{0}) = a s$ , so $v^{2} = u^{2} + 2 a s$	1

(a) Draw the v-t graph of a body moving with uniform acceleration a, starting with velocity u. (b) Use it to derive $v = u + a t$ and $s = u t + \frac{1}{2} a t^{2}$ . (c) A car moving at 54 km h $^{- 1}$ needs 0.4 s to react and then decelerates at 5 m s $^{- 2}$ . Find the total stopping distance. [5 marks] (5 marks)

Step	Marks
(a) Straight line from (0, u) to (t, v) with positive slope, labelled axes	1
(b) Slope $= a = \frac{v - u}{t} \Rightarrow v = u + a t$	1
Area = rectangle $u t$ + triangle $\frac{1}{2} (v - u) t = \frac{1}{2} a t^{2}$ , so $s = u t + \frac{1}{2} a t^{2}$	1
(c) u = 15 m s $^{- 1}$ ; reaction distance $= 15 \times 0.4 = 6 m$	1
Braking distance $= \frac{1 5 ^{2}}{2 \times 5} = 22.5 m$ ; total = 28.5 m	1

Where students lose marks in Motion in a Straight Line

The slips examiners see most often, and what to write instead.

Mistake: Mixing up sign conventions in vertical motion
Taking g positive when up is positive.
Wrong working
Ball thrown up at 10 m s $^{- 1}$ : $s = 10 t + 5 t^{2}$
Right working
Up positive: $s = 10 t - 5 t^{2}$ (g = 10)
Do this: State your positive direction at the start and give every vector a sign.
Usually costs: Whole numerical.
Mistake: Using the equations of motion when acceleration is not constant
Habit.
Do this: If a depends on t or x, integrate instead: $v = \int a d t$ , $x = \int v d t$ .
Usually costs: 2-3 marks.
Mistake: Reporting displacement as distance
Not checking whether the particle turns back.
Wrong working
$x = t^{3} - 3 t^{2}$ : distance in 3 s = x(3) − x(0) = 0 m
Right working
v = 0 at t = 2 s; x(0) = 0, x(2) = −4 m, x(3) = 0, so distance = 4 + 4 = 8 m
Do this: Find when v = 0, split the interval there, and add the sizes of each part.
Usually costs: 1-2 marks.
Mistake: Taking average speed as the mean of two speeds
Ignoring that time spent differs.
Do this: Always use total distance ÷ total time; for equal distances this is $\frac{2 v _{1} v _{2}}{v _{1} + v _{2}}$ .
Usually costs: 1 mark.
Mistake: Forgetting to convert km h $^{- 1}$ to m s $^{- 1}$
Rushing.
Do this: Multiply by $\frac{5}{18}$ : 72 km h $^{- 1}$ = 20 m s $^{- 1}$ , 54 km h $^{- 1}$ = 15 m s $^{- 1}$ .
Usually costs: 1 mark.
Mistake: Drawing impossible graphs
Not thinking about what the graph means.
Do this: An x-t graph cannot turn back in time or be vertical; a speed-time graph cannot go below zero; path length never decreases.
Usually costs: 1 mark in graph questions.

NCERT exercise guide: Motion in a Straight Line

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 2.3-2.6 (numericals with constant acceleration)

What it tests: Average speed for trips, stopping a vehicle, and a ball thrown up with g as the only acceleration.

How to approach it: Convert units, choose a positive direction, list u, v, a, s, t and pick the equation without the unknown you do not need.

Same kind as Exercise 2.6

A ball is thrown straight up with a speed of 19.6 m s $^{- 1}$ . Taking g = 9.8 m s $^{- 2}$ , find (a) the maximum height, (b) the time to return to the hand, (c) its velocity and acceleration at the highest point.

Show the solutionSolution

(a) $H = \frac{u ^{2}}{2 g} = \frac{19. 6 ^{2}}{19.6} = 19.6 m$
(b) $T = \frac{2 u}{g} = \frac{39.2}{9.8} = 4 s$
(c) At the top v = 0 but a = 9.8 m s $^{- 2}$ downwards (gravity never switches off).

Answer: (a) 19.6 m (b) 4 s (c) v = 0, a = 9.8 m s $^{- 2}$ down

Exercises 2.7 and 2.9 (concept statements)

What it tests: True/false statements on speed, velocity and acceleration, and the difference between quantities in an interval and at an instant.

How to approach it: Test each statement against an example: a ball at the top of its flight, a car going round a U-turn, a round trip.

Same kind as Exercise 2.7

True or false, with a reason: 'A particle whose velocity is negative must be slowing down.'

Show the solutionSolution

Whether it slows down depends on the signs of v and a together, not on the sign of v alone.
Example: with up as positive, a dropped stone has negative velocity and negative acceleration (−g), and its speed keeps increasing.

Answer: False: it slows down only if the acceleration is positive (opposite to v).

Exercises 2.12, 2.13 and 2.15-2.18 (graphs)

What it tests: Identifying impossible graphs, reading x-t and speed-time graphs, signs of velocity and acceleration in intervals, and finding distance from areas.

How to approach it: Slopes and areas; a body cannot be at two positions at once, speed cannot be negative, path length cannot decrease.

Same kind as Exercise 2.18

A body starts from rest, accelerates uniformly to 12 m s $^{- 1}$ in 4 s, moves at that speed for 6 s and then slows uniformly to rest in 3 s. Find the total distance and the average speed.

Show the solutionSolution

Distance = area under v-t graph = $\frac{1}{2} (4) (12) + (12) (6) + \frac{1}{2} (3) (12)$
= 24 + 72 + 18 = 114 m
Average speed $= \frac{114}{13} \approx 8.8 m s^{- 1}$

Answer: 114 m; about 8.8 m s $^{- 1}$

Practice questions on Motion in a Straight Line

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The area under a velocity-time graph gives:
(a) acceleration
(b) displacement
(c) speed
(d) jerk
Hint
Velocity × time.
Answer
(b) displacement
Q2 · 1 mark · MCQExam level
A body falls freely from rest. The distances it covers in the 1st, 2nd and 3rd seconds are in the ratio:
(a) 1 : 2 : 3
(b) 1 : 4 : 9
(c) 1 : 3 : 5
(d) 1 : 1 : 1
Hint
Galileo's law.
Answer
(c) 1 : 3 : 5
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): A position-time graph can never be a vertical line.
Reason (R): A body cannot be at two different positions at the same instant.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
A vertical line means many x at one t.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Can the average speed of a body be zero while it is moving? Can its average velocity be zero? Give one example.
Hint
Round trip.
Answer
Average speed cannot be zero if the body has moved (path length > 0). Average velocity can be zero: a runner who completes one lap and returns to the start has zero displacement.
Q5 · 2 marks · Very short answerHarder exam level
The velocity of a particle is $v = 3 t^{2} - 2 t$ (m s $^{- 1}$ ). Find its displacement between t = 0 and t = 2 s.
Hint
Integrate v.
Answer
$s = \int_{0}^{2} (3 t^{2} - 2 t) d t = [t^{3} - t^{2}]_{0}^{2} = 8 - 4 = 4 m$ .
Q6 · 2 marks · Very short answerHarder exam level
A car starts from rest with acceleration 2 m s $^{- 2}$ . How far does it go in the 5th second?
Hint
$s_{n th} = u + \frac{a}{2} (2 n - 1)$ .
Answer
$s_{5} = 0 + 1 \times 9 = 9 m$ (check: 25 m in 5 s minus 16 m in 4 s).
Q7 · 2 marks · Very short answerHarder exam level
A train increases its speed uniformly from 36 km h $^{- 1}$ to 72 km h $^{- 1}$ in 20 s. Find its acceleration and the distance covered in this time.
Hint
10 m s $^{- 1}$ to 20 m s $^{- 1}$ .
Answer
$a = \frac{20 - 10}{20} = 0.5 m s^{- 2}$ ; $s = \frac{10 + 20}{2} \times 20 = 300 m$ .
Q8 · 3 marks · Short answerHarder exam level
The velocity of a particle depends on its position as $v = 4 x$ (SI units). Show that its acceleration is constant and find it.
Hint
$a = v d v / d x$ .
Answer
$\frac{d v}{d x} = \frac{2}{x}$ , so $a = v \frac{d v}{d x} = 4 x \times \frac{2}{x} = 8 m s^{- 2}$ , independent of x: constant.
Q9 · 3 marks · Short answerHarder exam level
Sketch the x-t, v-t and a-t graphs for a ball thrown vertically up and caught at the same height (take up as positive). Explain the shape of each.
Hint
Parabola, straight line, flat line.
Answer
x-t: an inverted parabola, rising to a maximum at $t = u / g$ and returning to zero at $2 u / g$ . v-t: a straight line falling from +u through zero (at the top) to −u, slope −g. a-t: a horizontal line at −g throughout, since gravity acts all the time.
Q10 · 3 marks · Short answerHOTS / competency
The position of a particle is $x = t^{3} - 6 t^{2} + 9 t$ (x in m, t in s). Find its displacement and the distance travelled in the first 4 s.
Hint
Find when v = 0.
Answer
$v = 3 t^{2} - 12 t + 9 = 3 (t - 1) (t - 3)$ , zero at t = 1 s and 3 s. x(0) = 0, x(1) = 4, x(3) = 0, x(4) = 4 m. Displacement = 4 m; distance = 4 + 4 + 4 = 12 m.
Q11 · 2 marks · Very short answerHOTS / competency
A car's braking distance at 60 km h $^{- 1}$ is 20 m. With the same retardation, what is it at 120 km h $^{- 1}$ ? Explain why speed limits near schools are so effective.
Hint
$d_{s} \propto u^{2}$ .
Answer
Doubling speed multiplies braking distance by 4: 80 m. Because stopping distance grows with the square of speed, a small cut in speed gives a large cut in stopping distance.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define instantaneous velocity.Show answer
The limit of average velocity as the time interval tends to zero: $v = d x / d t$ , the slope of the tangent to the x-t graph.
2.What does the area under a v-t graph give?Show answer
The displacement in that interval.
3.Can a body have zero velocity but non-zero acceleration?Show answer
Yes: a ball at the top of its upward flight has v = 0 but a = g downwards.
4.Write the three equations of uniformly accelerated motion.Show answer
$v = u + a t$ , $s = u t + \frac{1}{2} a t^{2}$ , $v^{2} = u^{2} + 2 a s$ .
5.How does stopping distance depend on initial speed?Show answer
$d_{s} = u^{2} /2 a$ , so it is proportional to the square of the speed.
6.State Galileo's law of odd numbers.Show answer
For a body falling from rest, distances in successive equal time intervals are in the ratio 1 : 3 : 5 : 7 …
7.Write acceleration in terms of velocity and position.Show answer
$a = v d v / d x$ .
8.When is average speed equal to the magnitude of average velocity?Show answer
When the object moves in one direction along a straight line without turning back.

Best order to study it

Set up a frame of reference; learn displacement vs path length and uniform vs non-uniform motion.
Learn average velocity and speed and read x-t graphs.
Learn instantaneous velocity and acceleration as derivatives, with the power rule.
Learn what slopes and areas of x-t and v-t graphs mean.
Derive the three equations from the v-t graph, then by integration.
Practise free fall, vertical throws, stopping distance and reaction time, and calculus problems with changing acceleration.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain a frame of reference with an example?
Can you show when average speed differs from average velocity?
Can you find v and a from a given x(t)?
Can you find v(t) and x(t) from a given a(t) and initial values?
Can you read displacement and acceleration from a v-t graph?
Can you derive all three equations of motion from a v-t graph?
Can you derive $v^{2} = u^{2} + 2 a s$ using $a = v d v / d x$ ?
Can you solve a 'ball thrown up from a tower' problem with signs?
Can you compute total stopping distance including reaction time?
Can you sketch x-t, v-t and a-t graphs for a ball thrown up and caught?

Motion in a Straight Line: frequently asked questions

Is relative velocity in the Class 11 syllabus for 2026-27?

No. Relative velocity in one and two dimensions was removed in the NCERT rationalisation and is not in the CBSE content for 2026-27. Problems like two balls meeting can still be solved by writing each one's position equation.

Do I need calculus for this chapter?

Yes, at an elementary level. The syllabus names elementary differentiation and integration for describing motion, and the relations for uniformly accelerated motion by graphical and calculus treatment. You need the power rule for derivatives and simple integrals of polynomials.

Which derivation of the equations of motion should I learn?

Both. The syllabus asks for graphical and calculus treatment, so be ready to derive each equation from the v-t graph (slope and areas) and by integration.

Why is acceleration due to gravity taken negative sometimes?

Its sign depends only on your chosen positive direction. If up is positive, g points down, so a = −g. If down is positive, a = +g. Both give the same physical answer.

How is this chapter tested in Class 11?

In the school annual exam, through MCQs and assertion-reason items on graphs and definitions, short numericals on free fall and stopping distance, and 3-5 mark derivations of the equations of motion. JEE and NEET test the calculus and graph ideas heavily.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part I, Textbook for Class XI, Chapter 2 (Reprint 2026-27)