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?

Category: QuestionsA 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?
Editor">Editor Staff asked 11 months ago

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√b
 
The 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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience

1 Answers
Editor">Editor Staff answered 11 months ago

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.