Kinematics (s, v, a) practice questions.

A particle moves in a straight line with displacement s(t) = t³ − 6t² + 9t metres, t ≥ 0. Determine the times at which the particle is instantaneously at rest.

• t = 2 only
• t = 1 and t = 3
• t = 0 and t = 3
• t = 3 only

Answer: t = 1 and t = 3

The particle is at rest when v(t) = s′(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3) = 0, giving t = 1 and t = 3.

A particle moves along a line with velocity v(t) = t² − 4t + 3 (m/s). Calculate its acceleration at t = 3 s.

• 0 m/s²
• 6 m/s²
• 2 m/s²
• −1 m/s²

Answer: 2 m/s²

Acceleration is the derivative of velocity: a(t) = 2t − 4, so a(3) = 2 m/s².

A particle moves in a straight line with velocity v(t) = 4 − 2t (m/s) for 0 ≤ t ≤ 4. Determine the total distance it travels in this interval.

• 0 m
• 4 m
• 16 m
• 8 m

Answer: 8 m

v changes sign at t = 2, so distance = ∫₀² (4 − 2t) dt + |∫₂⁴ (4 − 2t) dt| = 4 + 4 = 8 m, while the displacement is 0.

A particle has acceleration a(t) = 6t (m/s²) and initial velocity v(0) = −3 m/s. Determine v(t).

• v(t) = 3t² − 3
• v(t) = 6t² − 3
• v(t) = 3t² + 3
• v(t) = t³ − 3

Answer: v(t) = 3t² − 3

Integrate acceleration: v(t) = ∫6t dt = 3t² + c, and v(0) = −3 gives c = −3, so v(t) = 3t² − 3.

At a certain instant, a particle moving in a straight line has velocity and acceleration with opposite signs. State which conclusion is correct.

• The particle's speed is increasing
• The particle's speed is decreasing
• The particle is momentarily at rest
• The particle's displacement is zero

Answer: The particle's speed is decreasing

When v and a have opposite signs the acceleration acts against the motion, so the particle is slowing down (speed decreasing).

The displacement of a particle is s(t) = 2t³ − 3t² metres. Calculate its acceleration at t = 1 s.

• 12 m/s²
• 0 m/s²
• −6 m/s²
• 6 m/s²

Answer: 6 m/s²

Differentiate twice: v(t) = 6t² − 6t and a(t) = 12t − 6, so a(1) = 6 m/s².