Question

14(a) A particle initially at rest moves in a straight line with an acceleration of (10t−4t²)ms⁻². Find the: (i) velocity of the particle after t seconds; (ii) average velocity of the particle during the 4th second.
(b) A load of mass 120 kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards: (i) at a constant velocity; (ii) with an acceleration of 3ms⁻². [Take g=10ms⁻²]

Accepted Answer

(a)(i) v=∫(10t−4t²)dt = 5t²−(4/3)t³+k. Since v=0 at t=0, k=0, so v = 5t²−(4/3)t³.
(ii) Distance s=∫v dt = (5/3)t³−(1/4)... integrating gives s=(5/3)t³−(1/3)t⁴. s(4)=64/3≈21.33m, s(3)=18m. Average velocity during the 4th second = s(4)−s(3) = 64/3−18 ≈ 3.33 ms⁻¹.
(b)(i) At constant velocity (a=0): Reaction R = mg = 120×10 = 1200N.
(ii) With acceleration 3ms⁻²: using R=m(g−a) as applied in the original solution, R = 120(10−3) = 840N (note: if the lift accelerates upward with increasing speed, the standard formula R=m(g+a) would instead give 1560N — the direction of the given acceleration should be checked against the scenario).

WAEC Further Mathematics 2022 Theory — Question 14

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