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A particle starts moving from rest with an acceleration in a fixed direction. If its acceleration at time t be(a – bt^2),where a and b are positive constants then which one is correct?(a) [v]max = 4a√a/3√b(b) [v]max = 2a√a/3√b(c) [v]max = 2a√a/3√b(d) [v]max = 4a√a/3√bThe question was asked by my college director while I was bunking the class.This interesting question is from Calculus Application in portion Application of Calculus of Mathematics – Class 12

Answer»

Correct answer is (b) [v]max = 2a√a/3√b

The explanation: If v be the velocity of the moving PARTICLE at TIME t then its acceleration at time t will be dv/DT. By question,

dv/dt = a – bt^2

Integrating we GET, v = ∫ a – bt^2 dt = at – bt^3/3 + k……….(1)

where k is constant of integration.

Given, v = 0, when t = 0; hence from (1) we get,

0 = a(0) – b/3(0) + k

Or k = 0

Thus, v = at – bt^3/3……….(2)

Again, d^2v/dt^2 = d(a – bt^2)/dt = -2bt

Now, for minimum or maximum value of v we have,

dv/dt = 0

Or a – bt^2 = 0

Or t^2 = a/b

Or t = √a/√b[Since t > 0 and a, b are positive constants]

At t = √a/√b we have d^2v/dt^2 = -2b(√a/√b) < 0 [Since t < 0 and a, b are positive constants]

Putting t = √a/√b in (2),

We find, v is maximum at t = √a/√b and the minimum value of v is,

[v]max = 2a√a/3√b.


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