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Quiz Chapter 8: Kinematics of Linear Motion

10 questions · Form 5 Additional Mathematics Bab 8: Kinematics of Linear Motion

Question 1 of 10Score: 0

A car decelerates from 20 ms⁻¹ to rest in 5 seconds with constant deceleration. Find its acceleration.

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. A car decelerates from 20 ms⁻¹ to rest in 5 seconds with constant deceleration. Find its acceleration.

  1. -4 ms⁻²
  2. 4 ms⁻²
  3. -5 ms⁻²
  4. 0 ms⁻²
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Answer: A

a = v - ut = 0 - 205 = -4 ms⁻².

2. What does a negative acceleration (a < 0) represent?

  1. Deceleration or retardation
  2. Increasing speed
  3. Instantaneous rest
  4. Zero displacement
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Answer: A

Negative acceleration indicates that velocity is decreasing with respect to time, which is deceleration.

3. What is the difference between displacement and total distance?

  1. Displacement is vector (can be negative) representing net position; distance is scalar (always ≥ 0) representing total path length
  2. Displacement is always greater than distance
  3. Distance depends on direction, displacement does not
  4. Displacement is measured in ms⁻¹, distance in meters
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Answer: A

Displacement measures net change in position (vector), whereas total distance sums up all physical movement along the path (scalar).

4. The acceleration of an object is given by a = 6t. If initial velocity is 0, find displacement s as a function of t (assuming s = 0 at t = 0).

  1. s = t³
  2. s = 3t²
  3. s = 6t³
  4. s = t³ / 3
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Answer: A

v = ∫ 6t dt = 3t² + c (since v(0)=0, c=0). s = ∫ 3t² dt = t³ + d (since s(0)=0, d=0). Thus, s = t³.

5. A particle moves along a straight line such that s = t³ - 3t. Find the times when the particle returns to the fixed point O.

  1. t = 0 s and t = √3 s
  2. t = 1 s and t = 3 s
  3. t = 0 s and t = 3 s
  4. t = √3 s only
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Answer: A

At fixed point O, s = 0 => t³ - 3t = 0 => t(t² - 3) = 0. Since time t ≥ 0, t = 0 s and t = √3 s.

6. Given v = 3t² - 12, find the displacement during the time interval 0 ≤ t ≤ 2 s.

  1. -16 m
  2. 16 m
  3. -24 m
  4. 8 m
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Answer: A

s = ∫₀² (3t² - 12) dt = [t³ - 12t]₀² = (8 - 24) - 0 = -16 m.

7. Given acceleration a = 6 ms⁻². If initial velocity is 2 ms⁻¹, find the velocity after 4 seconds.

  1. 26 ms⁻¹
  2. 24 ms⁻¹
  3. 20 ms⁻¹
  4. 12 ms⁻¹
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Answer: A

Since a is constant, v = u + at = 2 + (6)(4) = 2 + 24 = 26 ms⁻¹.

8. An object passes through fixed point O when t = 0. What is its displacement at t = 0?

  1. s = 0
  2. s = v
  3. s = 1
  4. s = a
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Answer: A

By definition, the fixed reference point O has displacement s = 0.

9. What is the condition for a particle to be instantaneously at rest?

  1. v = 0
  2. s = 0
  3. a = 0
  4. t = 0
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Answer: A

A particle is at rest when its velocity v is equal to 0.

10. If velocity v = 6t - 18, at what time t does the particle reverse its direction of motion?

  1. t = 3 s
  2. t = 6 s
  3. t = 18 s
  4. t = 0 s
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Answer: A

Direction reverses when v = 0. Set 6t - 18 = 0 => 6t = 18 => t = 3 s.

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