§ 2 ✓NCERT Class 11 · Physics · Chapter 2

Chapter test: Motion in a Straight Line

30 questions · Solve each question on paper first, then open "Show answer" to check your working.

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026

Q1 · NUMERICAL VALUE · DISTANCE & DISPLACEMENT
A particle moves 8 m towards east and then 3 m towards west. The distance travelled is ______ m.
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11 — distance adds path with no signs: 8 + 3 = 11 m (displacement would be +5 m)
Q2 · NUMERICAL VALUE · DIFFERENTIATION
A body's position is x = 5t² + 3t (SI units). Its speed at t = 2 s is ______ m s⁻¹.
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23 — v = dx/dt = 10t + 3 → at t = 2: 23 m s⁻¹
Q3 · MCQ · GRAPHS
On a position–time graph, the slope gives:
1. displacement
2. velocity
3. acceleration
4. distance
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(b) — slope of x–t = dx/dt = velocity (chord → average; tangent → instantaneous)
Q4 · NUMERICAL VALUE · AVERAGE SPEED
A car covers two equal stretches of road at 40 km h⁻¹ and 60 km h⁻¹. Its average speed for the whole journey is ______ km h⁻¹.
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48 — equal distances → harmonic mean 2(40)(60)/100 = 48 km h⁻¹, not 50
Q5 · NUMERICAL VALUE · FREE FALL
A ball dropped from rest falls 19.6 m before hitting the ground. Taking g = 9.8 m s⁻², the time of fall is ______ s.
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2 — h = ½gt² → 19.6 = 4.9t² → t = 2 s
Q6 · NUMERICAL VALUE · BRAKING
A car moving at 20 m s⁻¹ brakes uniformly at 5 m s⁻². The distance it covers while stopping is ______ m.
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40 — v² = v₀² + 2aΔx → 0 = 400 − 10Δx → Δx = 40 m
Q7 · NUMERICAL VALUE · NTH SECOND
A body starts from rest with a = 2 m s⁻². The distance it covers in the 3rd second is ______ m.
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5 — sₙ = v₀ + a(2n−1)/2 = 0 + 2(5)/2 = 5 m
Q8 · MCQ · FREE FALL
A ball thrown vertically upward is at its highest point. At that instant:
1. v = 0, a = 0
2. v = 0, a = 9.8 m s⁻² upward
3. v = 0, a = 9.8 m s⁻² downward
4. v = 9.8 m s⁻¹ upward, a = 0
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(c) — gravity never pauses: v = 0 momentarily, a = g = 9.8 m s⁻² downward
Q9 · NUMERICAL VALUE · RELATIVE VELOCITY
Two trains run toward each other on parallel tracks at 54 km h⁻¹ and 36 km h⁻¹. The speed of one as seen from the other is ______ m s⁻¹.
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25 — 54 → 15 m/s, 36 → 10 m/s; head-on → 15 + 10 = 25 m s⁻¹
Q10 · MCQ · GRAPH INTERPRETATION
From the velocity–time graph below, the total displacement over the 8 s journey is:

Figure T-10 — v–t record for Q10

Read the area between the line and the t-axis: a 2 s ramp, a 4 s plateau at 10 m s⁻¹, and a 2 s ramp down.

v (m s⁻¹) t (s)

10 0 2 6 8
1. 50 m
2. 60 m
3. 70 m
4. 80 m
Show answer
(b) — areas: ½(2)(10) + (4)(10) + ½(2)(10) = 10 + 40 + 10 = 60 m
Q11 · MCQ · KINEMATIC EQUATIONS
A car accelerates uniformly from 10 m s⁻¹ to 30 m s⁻¹ over a 200 m stretch. Its acceleration is:
1. 1 m s⁻²
2. 2 m s⁻²
3. 4 m s⁻²
4. 5 m s⁻²
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(b) — v² = v₀² + 2aΔx → 900 = 100 + 400a → a = 2 m s⁻²
Q12 · MCQ · AVERAGES
For motion along a straight line, the average speed and the magnitude of the average velocity obey:
1. average speed ≥ |average velocity|, always
2. average speed ≤ |average velocity|, always
3. they are always equal
4. no relation exists between them
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(a) — divide d ≥ |s| by Δt; equality only with no reversal
Q13 · MCQ · RELATIVE VELOCITY
A moves at +15 m s⁻¹ and B at −5 m s⁻¹ along the same line. The velocity of A relative to B is:
1. 10 m s⁻¹
2. 20 m s⁻¹
3. 15 m s⁻¹
4. 5 m s⁻¹
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(b) — v_AB = v_A − v_B = 15 − (−5) = 20 m s⁻¹ (signs do the work)
Q14 · MCQ · VALIDITY
Which of the three standard kinematic equations remain valid when the acceleration changes with time?
1. only v = v₀ + at
2. only v² = v₀² + 2aΔx
3. none of them
4. all of them
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(c) — all three derive from ∫a dt with a constant; variable a voids every one
Q15 · NUMERICAL VALUE · GRAPHS
A body's velocity grows uniformly from 0 to 20 m s⁻¹ in 4 s. Its displacement in this time is ______ m.
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40 — triangle area ½ × 4 × 20 = 40 m (or x = ½at² with a = 5)
Q16 · MCQ · GRAPHS
A position–time graph is a horizontal straight line over an interval. The particle is:
1. at rest throughout
2. moving with constant velocity
3. moving with constant acceleration
4. reversing repeatedly
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(a) — zero slope at every instant → v = 0 throughout → rest
Q17 · ASSERTION–REASON
Assertion (A): A heavy stone and a light stone dropped together from the same height (no air resistance) hit the ground at the same instant.
Reason (R): The acceleration of free fall, g, is the same for all bodies at a given place.
Options: (a) both true, R explains A · (b) both true, R does not explain A · (c) A true, R false · (d) A false, R true
Show answer
(a) — g's mass-independence is exactly why both stones share t = √(2h/g)
Q18 · STATEMENT
Statement I: A particle can have zero displacement while covering a non-zero distance.
Statement II: Distance is a scalar and displacement a vector.
Options: (a) both true, II explains I · (b) both true, II does not explain I · (c) I correct, II incorrect · (d) I incorrect, II correct
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(b) — both true (round trips; scalar-vs-vector), but II is a labelling fact, not the cause of I
Q19 · ASSERTION–REASON
Assertion (A): Negative acceleration always means the body is slowing down.
Reason (R): A body slows down when its acceleration is opposite to its velocity.
Options: (a) both true, R explains A · (b) both true, R does not explain A · (c) A true, R false · (d) A false, R true
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(d) — A false (v·a > 0 can still hold); R is the correct slowing condition
Q20 · STATEMENT
Statement I: The equation v = v₀ + at is applicable only when the acceleration is constant.
Statement II: The equation follows from integrating a = dv/dt with a treated as constant.
Options: (a) both true, II explains I · (b) both true, II does not explain I · (c) I correct, II incorrect · (d) I incorrect, II correct
Show answer
(a) — integrating dv = a dt with constant a yields v = v₀ + at — II is the derivation
B1 · MCQ
A particle's position is x = 3t³ − 6t² (SI). It is momentarily at rest at t =
1. 2/3 s
2. 1 s
3. 4/3 s
4. 2 s
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(c) 4/3 s — v = 9t² − 12t = 3t(3t − 4) = 0 → t = 4/3 s (t = 0 is the start, not a reversal)
B2 · NUMERICAL VALUE
A body accelerates uniformly from rest to 20 m s⁻¹ in 5 s and then continues at 20 m s⁻¹ for another 5 s. Its total displacement is ______ m.
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150 — ramp triangle ½ × 5 × 20 = 50 m; plateau 20 × 5 = 100 m; total 150 m
B3 · MCQ
Two cars 500 m apart drive toward each other at 20 m s⁻¹ and 30 m s⁻¹. They meet after:
1. 5 s
2. 10 s
3. 12.5 s
4. 25 s
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(b) 10 s — head-on approach 20 + 30 = 50 m s⁻¹; t = 500/50 = 10 s
B4 · MCQ
A body starts from rest with uniform acceleration. The ratio of distances covered in the 1st, 2nd and 3rd seconds is:
1. 1 : 2 : 3
2. 1 : 3 : 5
3. 1 : 4 : 9
4. 1 : 3 : 9
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(b) 1 : 3 : 5 — sₙ ∝ (2n − 1) from rest; totals 1, 4, 9 → differences 1, 3, 5
B5 · MCQ
A ball is thrown vertically upward at 19.6 m s⁻¹ (g = 9.8 m s⁻²). Its total time of flight back to the thrower's hand is:
1. 2 s
2. 3 s
3. 4 s
4. 8 s
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(c) 4 s — t_up = u/g = 2 s; T = 2u/g = 4 s (symmetry, no drag)
B6 · MCQ
A ball's velocity changes from +10 m s⁻¹ to −10 m s⁻¹ in 4 s along a line. The magnitude of its average acceleration is:
1. 0
2. 5 m s⁻²
3. 10 m s⁻²
4. 20 m s⁻²
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(b) 5 m s⁻² — Δv = (−10) − (+10) = −20 m s⁻¹; |ā| = 20/4 = 5 m s⁻² — reversal doubles Δv
B7 · MCQ
For the same braking acceleration, if a car's speed is made three times larger, its stopping distance becomes:
1. the same
2. 3 times
3. 6 times
4. 9 times
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(d) 9 times — d = v²/2|a| with |a| fixed → d ∝ v²; ×3 speed → ×9 distance
B8 · MCQ
An object moves from x = 0 to x = 10 m in 4 s, rests 2 s, and returns to x = 0 in the next 4 s. Its total distance and net displacement are:
1. 10 m and 0
2. 20 m and 10 m
3. 20 m and 0
4. 0 and 0
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(c) 20 m and 0 — distance = 10 + 10 = 20 m; ends where it began → displacement 0
B9 · NUMERICAL VALUE
A body starts from rest with a = 4 m s⁻². The distance covered in 5 s is ______ m.
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50 — x = ½at² = ½ × 4 × 25 = 50 m
B10 · NUMERICAL VALUE
A man stands 8 m behind a bus that starts from rest with a = 2 m s⁻². The man runs at a steady 6 m s⁻¹. He catches the bus at t = ______ s.
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2 — 6t = 8 + ½(2)t² → t² − 6t + 8 = 0 → t = 2 s (first root; he boards then)

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Figure T-10 — v–t record for Q10