Correct option is (b) 22(2/9)m
The best I can explain: Let f be the uniform retardation in m/sec^2 to the motion of the motor car due to application of brakes.
By question, the car is stopped by its brakes in 4 seconds, hence, the final velocity of the car after 4 seconds = 0.
Therefore, using the formula v = u – ft we get,
0 = ((40*1000)/(60*60) – f(4)) [Since u = initial velocity of the motor car = 40 km/hr = (40*1000)/(60*60) m/sec]
Or f = 25/9
Let the car go through a distance s m from the point at which the brakes are first applied.
Then using the formula s = ut – 1/2(ft^2) we get,
s = ((40*1000)/(60*60))*4 – 1/2(25/9)(4*4)
= 200/9
= 22(2/9)
Therefore, the required distance described by the car = 22(2/9)m.