Question

A car A 1s travelling on a straight level road with a uniform speed of 60 kmph. it is followed by another car B which is moving with a speed of 70 kmph. When the distance between them is 2.5 km, the car B is given a deceleration of 20 kmh. The distance and time after which the car B catches up with the car, A are​

Answer 1

Answer:

32.5 km, 0.5 h

Explanation:

Answer 2

Explanation:

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

For car B: Using the second equation of motion: s = ut + 1 2 12at2

where: s = distance covered = s2

u = initial velocity = 70 km/h

a = acceleration = -20 km/h

t = time = t

s = ut + 1 2 12at2

s2 = 70t + 1 2 12 (-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