Kinematic problems (s, v, a) practice questions.

A particle moves in a straight line with displacement s = 2t³ − 5t² + 4t metres at time t seconds. Determine its velocity at t = 3 s.

• 19 m s⁻¹
• 28 m s⁻¹
• 9 m s⁻¹
• 34 m s⁻¹

Answer: 28 m s⁻¹

v = ds/dt = 6t² − 10t + 4; at t = 3, v = 54 − 30 + 4 = 28 m s⁻¹.

The velocity of a particle is v = 3t² − 12t + 9 m s⁻¹. Calculate its acceleration at t = 1 s.

• −6 m s⁻²
• 6 m s⁻²
• 0 m s⁻²
• −3 m s⁻²

Answer: −6 m s⁻²

a = dv/dt = 6t − 12; at t = 1, a = 6 − 12 = −6 m s⁻².

A particle moves with velocity v = t² − 5t + 6 m s⁻¹ for t ≥ 0. Determine the times at which it is momentarily at rest.

• t = 2 and t = 3
• t = 1 and t = 6
• t = −2 and t = −3
• t = 0 and t = 5

Answer: t = 2 and t = 3

Rest when v = 0: t² − 5t + 6 = (t − 2)(t − 3) = 0, so t = 2 s and t = 3 s.

A particle has velocity v = t² − 4t m s⁻¹ for 0 ≤ t ≤ 5. Calculate the total distance travelled (in metres).

• 13 m
• 5/3 m
• 32/3 m
• 8 m

Answer: 13 m

v = 0 at t = 0 and t = 4. Distance = |∫₀⁴ v dt| + |∫₄⁵ v dt| = |−32/3| + |7/3| = 32/3 + 7/3 = 13 m.

A particle's displacement is s = t³ − 6t² + 9t m. Determine the total distance travelled during 0 ≤ t ≤ 4.

• 4 m
• 8 m
• 20 m
• 12 m

Answer: 12 m

v = 3t² − 12t + 9 = 3(t−1)(t−3); s(0)=0, s(1)=4, s(3)=0, s(4)=4. Distance = 4 + 4 + 4 = 12 m.

A ball is thrown upward with velocity v = 20 − 9.8t m s⁻¹. Determine the maximum height reached (to 3 s.f.), taking initial height as 0.

• 20.4 m
• 40.8 m
• 10.2 m
• 2.04 m

Answer: 20.4 m

Max height when v = 0: t = 20/9.8 ≈ 2.041 s. Height = ∫₀²·⁰⁴¹ (20 − 9.8t) dt = 20(2.041) − 4.9(2.041)² ≈ 20.4 m.