Question

8. A force of F N acts on a body of mass 50 kg and changes its velocity from 30 ms⁻¹ to 40 ms⁻¹ in 6 seconds. Find: (a) the magnitude of the force; (b) the acceleration of the body; (c) the distance covered in 9 seconds when the initial velocity is 30 ms⁻¹. [Take g = 10 ms⁻²]

Accepted Answer

(a)&(b) a = (v-u)/t = (40-30)/6 = 5/3 ms⁻². Force F = ma = 50 × 5/3 = 250/3 ≈ 83.3 N.
(c) With u=30 ms⁻¹, a=5/3 ms⁻², t=9s: s = ut + ½at² = 30(9) + ½(5/3)(81) = 270+67.5 = 337.5 m.

WAEC Further Mathematics 2025 Theory — Question 8

Sign up free to unlock