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?
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)
Therefore, the required distance described by the car = 22(2/9)m.