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Motion in a Straight Line – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 2. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Motion along a line is described by position, displacement, velocity and acceleration. For uniform acceleration, three kinematic equations connect them, and graphs show motion visually.

Basic quantities

Displacement is the change in position (vector); distance is path length (scalar).
Average velocity = displacement/time; instantaneous velocity v = dx/dt; acceleration a = dv/dt.
Slope of x–t graph = velocity; slope of v–t graph = acceleration; area under v–t graph = displacement.

Uniform acceleration

v = u + at; s = ut + ½at²; v² = u² + 2as.
Distance in the nth second: sₙ = u + a(2n − 1)/2.
Free fall: a = g ≈ 9.8 m/s² downward; relative velocity v_AB = v_A − v_B.

Key terms

Displacement: shortest directed distance from initial to final position
Uniform acceleration: velocity changes by equal amounts in equal times
Relative velocity: velocity of one object as seen from another, v_AB = v_A − v_B

Formulas

v = u + at
s = ut + ½at²
v² = u² + 2as

Important questions with answers

Can a body have zero velocity and non-zero acceleration? Give an example.2 marks

Yes. A ball thrown vertically upward has zero velocity at the highest point, but its acceleration is still g downward.

Derive v² = u² + 2as from a velocity–time graph.3 marks

For uniform acceleration the v–t graph is a straight line from u to v in time t. Displacement s = area of the trapezium = ½(u + v)t. Also a = (v − u)/t, so t = (v − u)/a. Then s = ½(u + v)(v − u)/a = (v² − u²)/2a, giving v² = u² + 2as.

A car moving at 20 m/s applies brakes and stops in 50 m. Find the deceleration and the time taken. If the driver's reaction time is 0.5 s, find the total stopping distance from when the danger is seen.5 marks

v² = u² + 2as: 0 = 400 + 2a × 50, a = −4 m/s². t = (0 − 20)/(−4) = 5 s. During the reaction time the car moves 20 × 0.5 = 10 m at constant speed. Total stopping distance = 10 + 50 = 60 m.

Quick revision

Area under v–t = displacement.
Slope of x–t = velocity.
Free-fall acceleration does not depend on mass.

Exam tips

For a body thrown up, time up = time down = u/g and max height = u²/2g.
For a body thrown up, at the highest point v = 0 but a = g (downward) — acceleration is never zero there.

Practice MCQs

Slope of a velocity–time graph gives
1. acceleration
2. displacement
3. speed
4. distance
Show answer
Answer: (a) acceleration. dv/dt.
A car goes from 0 to 20 m/s in 5 s. Its acceleration is
1. 2 m/s²
2. 4 m/s²
3. 5 m/s²
4. 100 m/s²
Show answer
Answer: (b) 4 m/s². 20/5.
A body starts from rest with a = 2 m/s². Distance in 5 s is
1. 25 m
2. 10 m
3. 50 m
4. 20 m
Show answer
Answer: (a) 25 m. ½ × 2 × 25.
A ball is thrown up at 20 m/s (g = 10 m/s²). Maximum height is
1. 10 m
2. 40 m
3. 20 m
4. 2 m
Show answer
Answer: (c) 20 m. u²/2g.
A stone is dropped from a height of 80 m (g = 10 m/s²). Time to reach the ground is
1. 2 s
2. 4 s
3. 8 s
4. 16 s
Show answer
Answer: (b) 4 s. h = ½gt².
Distance covered in the 3rd second from rest with a = 4 m/s² is
1. 10 m
2. 18 m
3. 6 m
4. 12 m
Show answer
Answer: (a) 10 m. 0 + 4(5)/2.
Two trains move in the same direction at 60 km/h and 40 km/h. Relative speed is
1. 100 km/h
2. 50 km/h
3. 20 km/h
4. 2400 km/h
Show answer
Answer: (c) 20 km/h. Difference.
A ball thrown up returns to the thrower after 6 s (g = 10 m/s²). Its initial speed was
1. 60 m/s
2. 15 m/s
3. 30 m/s
4. 20 m/s
Show answer
Answer: (c) 30 m/s. Time up = 3 s.

All chapters of Class 11 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.