A motor car travelling at the rate of 40 km/hr is stopped by its brakes in 4 seconds. How long will it go from the point at which the brakes are first applied?

(a) 22m

(b) 22(2/9)m

(c) 22(1/9)m

(d) 22(4/9)m

This question was addressed to me in an international level competition.

This interesting question is from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

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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.