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Quiz Chapter 8: Kinematics of Linear Motion

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

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Full Question List & Answer Key

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

1. Given acceleration a = 4t - 8 and the initial velocity is 3 ms⁻¹, find the velocity function v(t).

  1. v = 2t² - 8t + 3
  2. v = 4t² - 8t + 3
  3. v = 2t² - 8
  4. v = 2t² - 8t
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Answer: A

v = ∫ (4t - 8) dt = 2t² - 8t + c. Since initial velocity v(0) = 3, c = 3. Thus, v = 2t² - 8t + 3.

2. Find the maximum velocity of a particle whose velocity function is v = 12t - 3t².

  1. 12 ms⁻¹
  2. 6 ms⁻¹
  3. 16 ms⁻¹
  4. 24 ms⁻¹
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Answer: A

Maximum velocity occurs when a = dv/dt = 0 => 12 - 6t = 0 => t = 2. Substitute t = 2 into v: v = 12(2) - 3(2)² = 24 - 12 = 12 ms⁻¹.

3. 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.

4. 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.

5. 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).

6. Given v = t² - 4t, find the displacement s in the first 3 seconds assuming s = 0 when t = 0.

  1. -9 m
  2. 9 m
  3. -3 m
  4. -18 m
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Answer: A

s = ∫ (t² - 4t) dt = [t³/3 - 2t²] from 0 to 3 = (273 - 2(9)) - 0 = 9 - 18 = -9 m.

7. 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.

8. Find the acceleration of a particle at t = 2 s if its velocity is v = 2t³ - 5t.

  1. 19 ms⁻²
  2. 24 ms⁻²
  3. 11 ms⁻²
  4. 16 ms⁻²
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Answer: A

a = dv/dt = 6t² - 5. At t = 2: a = 6(2)² - 5 = 24 - 5 = 19 ms⁻².

9. 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.

10. 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.

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