A car A is travelling on a straight level road with a uniform speed of 60 km/h. It is followed by another car B. Which is moving with a speed of 70 km/h. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km/h2. After how much time will B catch up with A?

Distance covered by car A in time t is, s1 = 60 t


For car B:


Using the second equation of motion: s = ut + at2


where: s = distance covered = s2
u = initial velocity = 70 km/h
a = acceleration = -20 km/h
t = time = t


s = ut + at2


s2 = 70t + (-20) × t2


s2 = 70t – 10 t2


But it is given that s2 – s1 =2.5 km
70t – 10 t2 - 60 t = 2.5


10t - 10 t2 = 2.5


t – t2 = 0.25


t – t2 – 0.25 =0


t2 –t +0.25 =0


(t – 0.5)2 =0


t = 0.5 hr

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