1
200 Multiple-Choice Questions on Motion in a Straight
Line (1D Kinematics) with Answers and Explanations
1: Displacement is best defined as:
A) Total ground covered
B) Change in position with direction
C) Speed with direction
D) Distance per unit time
Answer: B
Explanation: Displacement is a vector from initial to final position; distance is path length.
2: In 1D, velocity can be negative because:
A) Speed decreases
B) Time is negative
C) Motion is opposite the chosen positive direction
D) Acceleration is zero
Answer: C
Explanation: Sign encodes direction in 1D.
3: Average speed equals average velocity only when:
A) Acceleration is constant
B) Motion is uniform
C) Motion never reverses direction
D) Time is small
Answer: C
Explanation: If no reversal, distance = magnitude of displacement.
4:The SI unit of acceleration is:
A) m/s
B) m
C) m/s²
D) s/m
Answer: C
Explanation: a = Δv/Δt with v in m/s.
5: If x(t) = 5t − 2, velocity is:
A) 5t
B) 5
, 2
C) −2
D) 0
Answer: B
Explanation: v = dx/dt = 5 m/s (constant).
6: If v is constant and nonzero, acceleration is:
A) Constant and positive
B) Constant and negative
C) Zero
D) Variable
Answer: C
Explanation: a = dv/dt = 0.
7: The slope of a position–time graph gives:
A) Acceleration
B) Speed only
C) Velocity
D) Displacement
Answer: C
Explanation: Slope dx/dt equals velocity.
8: The area under a velocity–time curve gives:
A) Velocity
B) Acceleration
C) Displacement
D) Jerk
Answer: C
Explanation: ∫ v dt = Δx.
9: If velocity is always positive, the object:
A) Never accelerates
B) Never changes direction
C) Has zero displacement
D) Has zero speed
Answer: B
Explanation: Positive v means motion in + direction.
10: Acceleration is negative when:
A) Speed is negative
B) Velocity decreases in magnitude in + direction or increases in − direction
C) Time decreases
D) Displacement is negative
Answer: B
Explanation: Negative points opposite + axis.
11: For constant acceleration a, which is incorrect?
A) v = u + at
B) s = ut + ½at²
, 3
C) s = vt − ½at²
D) v² = u² + 2as
Answer: C
Explanation: s = vt − ½at² holds only if v is initial; correct is s = ut + ½at².
12: A car accelerates from rest at 2 m/s² for 5 s. Its speed is:
A) 2 m/s
B) 5 m/s
C) 10 m/s
D) 20 m/s
Answer: C
Explanation: v = 0 + 2×5 = 10 m/s.
13: Displacement in Q12 is:
A) 10 m
B) 25 m
C) 50 m
D) 100 m
Answer: C
Explanation: s = 0×5 + ½×2×25 = 25 m; wait: ½×2×25 = 25 m. Correct is 25 m.
Correction: Answer B (25 m)
Explanation: s = ½ a t² = ½×2×25 = 25 m.
14: If v(t) = 6 − 2t (m/s), the acceleration is:
A) 6 m/s²
B) −2 m/s²
C) 2 m/s²
D) 0
Answer: B
Explanation: a = dv/dt = −2.
15: For v(t) in Q14, the object stops at:
A) t = 0 s
B) t = 1 s
C) t = 3 s
D) t = 6 s
Answer: C
Explanation: Set v = 0 → 6 − 2t = 0 → t = 3 s.
16: If x(t) = t³, then acceleration at t = 2 s is:
A) 6 m/s²
B) 12 m/s²
C) 24 m/s²
D) 36 m/s²
Answer: C
Explanation: v = 3t², a = 6t → a(2) = 12; wait: derivative: x = t³ → v = 3t² → a = 6t. At t=2,
a=12. Correct answer B.
, 4
Answer: B
Explanation: a = d²x/dt² = 6t; at 2 s, 12 m/s².
17: Speed is:
A) Scalar magnitude of velocity
B) Vector quantity
C) Always negative
D) Same as acceleration
Answer: A
Explanation: Speed ≥ 0.
18: If average velocity over 10 s is 0, then:
A) Object was at rest
B) Displacement is zero
C) Distance is zero
D) Acceleration is zero
Answer: B
Explanation: Average velocity = Δx/Δt = 0 ⇒ Δx = 0; motion could be nonzero.
19: Which quantity can be negative?
A) Speed
B) Distance
C) Time
D) Displacement
Answer: D
Explanation: Displacement is signed.
20: The jerk is:
A) da/dt
B) dv/dt
C) dx/dt
D) d²x/dt²
Answer: A
Explanation: Jerk = rate of change of acceleration.
21: A car moves with v = 20 m/s for 15 s. Distance traveled:
A) 300 m
B) 150 m
C) 30 m
D) 3 m
Answer: A
Explanation: s = vt.
22: A stone dropped from rest falls 20 m. Speed just before impact (g = 10):
A) 10 m/s
B) 20 m/s
C) √400 m/s
D) √200 m/s
200 Multiple-Choice Questions on Motion in a Straight
Line (1D Kinematics) with Answers and Explanations
1: Displacement is best defined as:
A) Total ground covered
B) Change in position with direction
C) Speed with direction
D) Distance per unit time
Answer: B
Explanation: Displacement is a vector from initial to final position; distance is path length.
2: In 1D, velocity can be negative because:
A) Speed decreases
B) Time is negative
C) Motion is opposite the chosen positive direction
D) Acceleration is zero
Answer: C
Explanation: Sign encodes direction in 1D.
3: Average speed equals average velocity only when:
A) Acceleration is constant
B) Motion is uniform
C) Motion never reverses direction
D) Time is small
Answer: C
Explanation: If no reversal, distance = magnitude of displacement.
4:The SI unit of acceleration is:
A) m/s
B) m
C) m/s²
D) s/m
Answer: C
Explanation: a = Δv/Δt with v in m/s.
5: If x(t) = 5t − 2, velocity is:
A) 5t
B) 5
, 2
C) −2
D) 0
Answer: B
Explanation: v = dx/dt = 5 m/s (constant).
6: If v is constant and nonzero, acceleration is:
A) Constant and positive
B) Constant and negative
C) Zero
D) Variable
Answer: C
Explanation: a = dv/dt = 0.
7: The slope of a position–time graph gives:
A) Acceleration
B) Speed only
C) Velocity
D) Displacement
Answer: C
Explanation: Slope dx/dt equals velocity.
8: The area under a velocity–time curve gives:
A) Velocity
B) Acceleration
C) Displacement
D) Jerk
Answer: C
Explanation: ∫ v dt = Δx.
9: If velocity is always positive, the object:
A) Never accelerates
B) Never changes direction
C) Has zero displacement
D) Has zero speed
Answer: B
Explanation: Positive v means motion in + direction.
10: Acceleration is negative when:
A) Speed is negative
B) Velocity decreases in magnitude in + direction or increases in − direction
C) Time decreases
D) Displacement is negative
Answer: B
Explanation: Negative points opposite + axis.
11: For constant acceleration a, which is incorrect?
A) v = u + at
B) s = ut + ½at²
, 3
C) s = vt − ½at²
D) v² = u² + 2as
Answer: C
Explanation: s = vt − ½at² holds only if v is initial; correct is s = ut + ½at².
12: A car accelerates from rest at 2 m/s² for 5 s. Its speed is:
A) 2 m/s
B) 5 m/s
C) 10 m/s
D) 20 m/s
Answer: C
Explanation: v = 0 + 2×5 = 10 m/s.
13: Displacement in Q12 is:
A) 10 m
B) 25 m
C) 50 m
D) 100 m
Answer: C
Explanation: s = 0×5 + ½×2×25 = 25 m; wait: ½×2×25 = 25 m. Correct is 25 m.
Correction: Answer B (25 m)
Explanation: s = ½ a t² = ½×2×25 = 25 m.
14: If v(t) = 6 − 2t (m/s), the acceleration is:
A) 6 m/s²
B) −2 m/s²
C) 2 m/s²
D) 0
Answer: B
Explanation: a = dv/dt = −2.
15: For v(t) in Q14, the object stops at:
A) t = 0 s
B) t = 1 s
C) t = 3 s
D) t = 6 s
Answer: C
Explanation: Set v = 0 → 6 − 2t = 0 → t = 3 s.
16: If x(t) = t³, then acceleration at t = 2 s is:
A) 6 m/s²
B) 12 m/s²
C) 24 m/s²
D) 36 m/s²
Answer: C
Explanation: v = 3t², a = 6t → a(2) = 12; wait: derivative: x = t³ → v = 3t² → a = 6t. At t=2,
a=12. Correct answer B.
, 4
Answer: B
Explanation: a = d²x/dt² = 6t; at 2 s, 12 m/s².
17: Speed is:
A) Scalar magnitude of velocity
B) Vector quantity
C) Always negative
D) Same as acceleration
Answer: A
Explanation: Speed ≥ 0.
18: If average velocity over 10 s is 0, then:
A) Object was at rest
B) Displacement is zero
C) Distance is zero
D) Acceleration is zero
Answer: B
Explanation: Average velocity = Δx/Δt = 0 ⇒ Δx = 0; motion could be nonzero.
19: Which quantity can be negative?
A) Speed
B) Distance
C) Time
D) Displacement
Answer: D
Explanation: Displacement is signed.
20: The jerk is:
A) da/dt
B) dv/dt
C) dx/dt
D) d²x/dt²
Answer: A
Explanation: Jerk = rate of change of acceleration.
21: A car moves with v = 20 m/s for 15 s. Distance traveled:
A) 300 m
B) 150 m
C) 30 m
D) 3 m
Answer: A
Explanation: s = vt.
22: A stone dropped from rest falls 20 m. Speed just before impact (g = 10):
A) 10 m/s
B) 20 m/s
C) √400 m/s
D) √200 m/s
