A bus starts from rest and moves with constant acceleration 8 m/s². At the same time, a car travelling with a constant velocity 16 m/s overtakes and passes the bus. After how much time and at what distance, the bus overtakes the car?

The position of bus after time t is given as xb = (1/2)at². The position of car after time t is given as xc = ut. Hence when the bus passes the car, xb = xc ⇒ (1/2)at² = ut ⇒ t = 2u/a = 2 × 16 m/s / 8 m/s² = 4 s. Hence the distance at the crossing = ut = 16 m/s × 4 s = 64 m