Question

4. A car traveling on a long straight highway, at a
constant speed of 144 km/h passes a police
motorcycle moving at 72 km/h. Immediately after
the car passes the motorcycle, the police officer
accelerates the motorcycle at constant acceleration
of 2 m/s2 to overtake the speeding car. The police
motorcycle travels X metre to overtake the speeding
car. Then X =
(1) 100 m.
(2) 800 m
(3) 500 m
(4) None of these​

Answer 1

We consider the position where the car passes the police motorcycle and where the motorcycle is started to be accelerated as the initial position.

Here, according to the car,

u = 144 km h^(-1) = 40 m s^(-1)

a = 0 m s^(-2), because the car is in constant speed.

So we can say that,

X = 40 t → (1),

by the second kinematic equation,

s = u t + (a t²) / 2

And, according to the motorcycle,

u = 72 km h^(-1) = 20 m s^(-1)

a = 2 m s^(-2)

So,

X = 20 t + (2 t²) / 2

X = t² + 20 t → (2)

From (1) and (2),

t² + 20 t = 40 t

t² - 20 t = 0

t (t - 20) = 0

=> t = 0 ; t = 20

Well, t = 0 tells about the initial position, where the car meets the motorcycle. So we take t = 20 s.

Therefore, from (1),

X = 40 × 20 = 800 m

Hence (2) is the answer.