Question

A policeman is moving with constant speed on
a straight road. When he is at distance 250 m
behind a car, the car starts accelerating from rest
and move with a constant acceleration 2 m/s2. The
minimum speed of the policeman such that he can
catch the car is
(1) 10 m/s
(2) 1055 m/s
(3) 10/10 m/s (4) 10/2m/s

Answer 1

4 is the correct answer

Answer 2

Thus minimum value of v is around 31.62 m/s

Explanation:

Let t is the time duration taken by police to catch the thief.

 If a is acceleration, then distance travelled by thief staring from rest

= (1/2) at^2  = (1/2) × 2 × t^2 = t^2.........(1)  

If v is uniform speed, then distance travelled by police

d = v×t  .............(2)

 Since initially, Police is 200 m behind the thief, from eqn.(1) and (2), we get ,  v×t = t^2 + 250  ................(3)

 Hence  Equation.(3) is written as quadratic form as

t^2 - (v×t) + 250 = 0 ........(4)

Hence,   t = (1/2) [ v± ( v^2 - 1000 )^1/2 ]    ............  (5)

In equation (5), t is real only if  v^2 ≥ 1000    or   minimum value of v is around 31.62 m/s