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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

When an object moves with uniform velocity, what is 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. When an object moves with uniform velocity, what is its acceleration?

  1. a = 0
  2. a = v
  3. a > 0
  4. a = 9.8
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Answer: A

Uniform velocity means velocity is constant, so its derivative dv/dt = a = 0.

2. 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.

3. How is acceleration obtained from a displacement function s(t)?

  1. By differentiating s(t) twice with respect to t: d²s/dt²
  2. By integrating s(t) with respect to t
  3. By taking ds/dt
  4. By dividing s(t) by t²
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Answer: A

Velocity v = ds/dt and acceleration a = dv/dt = d²s/dt².

4. 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.

5. 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.

6. If displacement s = -15 m, what does this indicate about the position of the particle?

  1. It is 15 meters to the left (negative side) of fixed point O
  2. It is 15 meters to the right of fixed point O
  3. It has traveled a total distance of -15 meters
  4. It is moving backwards at 15 ms⁻¹
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Answer: A

Negative displacement means position is to the left or negative side of the reference origin O.

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. 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.

9. 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).

10. A particle moves such that its velocity is v = 4 - 2t. What is the total distance traveled from t = 0 to t = 3 s?

  1. 5 m
  2. 3 m
  3. 4 m
  4. 1 m
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Answer: A

Rest point v = 0 => 4 - 2t = 0 => t = 2 s. s = ∫ (4 - 2t) dt = 4t - t². At t = 0: s = 0. At t = 2: s = 4(2) - (2)² = 4 m. At t = 3: s = 4(3) - (3)² = 3 m. Distance = |4 - 0| + |3 - 4| = 4 + 1 = 5 m.

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