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)mThis 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

A motor car travelling at the rate of 40 km/hr is stopped by its brake

1.	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)mThis 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
Answer» 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 = ((401000)/(6060) – f(4))[Since u = initial velocity of the motor car = 40 km/hr = (401000)/(6060) 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 = ((401000)/(6060))4 – 1/2(25/9)(44) = 200/9 = 22(2/9) Therefore, the required distance described by the car = 22(2/9)m.

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