Additional Mathematics – Calculus | e-Consult
Calculus (1 questions)
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- Acceleration is the derivative of velocity: a(t) = dv/dt = –2 m s⁻². a = –2 m s⁻² (constant).
- Set v(t) = 0: 5 – 2t = 0 ⇒ t = 2.5 s. The particle stops at t = 2.5 s.
- Displacement from t = 0 to 2.5 s:
s₁ = ∫₀^{2.5}(5 – 2t) dt = [5t – t²]₀^{2.5} = (12.5 – 6.25) = 6.25 m.
From t = 2.5 s to 4 s the velocity is negative, so distance travelled is the magnitude of the displacement:
s₂ = ∫{2.5}^{4}(5 – 2t) dt = [5t – t²]{2.5}^{4}
= (20 – 16) – (12.5 – 6.25) = 4 – 6.25 = –2.25 m.
Distance for this interval = |–2.25| = 2.25 m.
Total distance = 6.25 m + 2.25 m = 8.50 m.