Most candidates found this question very difficult to solve. They only needed to understand the appropriate equation of motion to be applied. To find the initial velocity (u); since final velocity (v) = 55 m/s at time (t) = 5 secs and 105 m/s at time t = 10 secs, and using v = u + at, two simultaneous equations emerged, i.e 55 = u + 5q --- (i) and 105 = u + 10 a --- (ii) which when solved gave a = 10 m/s2.
The distance (s) = ut + ½ at 2 where u = 55 m/s, a = 10 m/s and t = (8-5) = 3 sec
i.e S = 55(3) + ½ (10)(3)2 = 210m.
