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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

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

Full Question List & Answer Key

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

1. 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.

2. 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.

3. 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.

4. Which mathematical operation is used to find displacement s from velocity v?

  1. Integration of v with respect to t
  2. Differentiation of v with respect to t
  3. Dividing v by time t
  4. Squaring velocity v
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Answer: A

Since v = ds/dt, integrating v gives displacement s = ∫ v dt.

5. 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.

6. 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.

7. 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.

8. 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.

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. An object has a displacement function s = t³ - 6t² + 9t. Find its initial velocity.

  1. 9 ms⁻¹
  2. 0 ms⁻¹
  3. -6 ms⁻¹
  4. 3 ms⁻¹
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Answer: A

Velocity v = ds/dt = 3t² - 12t + 9. Initial velocity occurs at t = 0: v = 3(0)² - 12(0) + 9 = 9 ms⁻¹.

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