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A particle moving in a straight line with uniform retardation described 7cm in 5th second and after some time comes to rest. If the particle describes 1/64 part of the total path during the last second of its motion, for how long was the particle in motion?(a) 6 seconds(b) 8 seconds(c) 4 seconds(d) 2 secondsI got this question in examination.This interesting question is from Calculus Application topic in chapter Application of Calculus of Mathematics – Class 12

Answer»

Correct answer is (b) 8 seconds

Explanation: Let the INITIAL velocity of the particle be u cm/second and its uniform retardation be f cm/sec^2.

Further ASSUME that the particle was in motion for t seconds.

By question, the particle comes to rest after t seconds.

Therefore, USING the formula, v = u – ft, we get,

0 = u – ft

Or u = ft

Again, the particle described 7cm in the 5th second. Therefore, using the formula

st = UT + 1/2(f)(2t – 1) we get,

7 = u – ½(f)(2.5 – 1)

Or u – 9f/2 = 7

Again, the distance described in the last second (i.e., in the t th second) of its motion

= 1/64 * (distance described by the particle in t seconds)

ut -1/2(f)(2t – 1) = 1/64(ut – ½(ft^2))

Putting u = ft, we get,

f/2 = 1/64((ft^2)/2)

Or t^2 = 64

Or t = 8 seconds.


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