We first find the lorry's initial velocity
a = 4ms−2, s= 250m, t= 10, u=? ⇒ ut + \(\frac{1}{2}×4×(10^2) ⇒250= 10u + 200 ⇒ 250-200 = 10u ⇒ 50 = 10u ⇒U =\frac{50}{10}\) = 5ms−1 For the next 10s, we set t=20s = s = (5 × 20) + 12×4×202 = s = 100 + 800
⇒ 900 m
;Distance covered:900-250 =650m
