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 seconds
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This interesting question is from Calculus Application topic in chapter Application of Calculus of Mathematics – Class 12
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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.