The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax^2 + bx + c (a, b, c are positive constants). If v be the velocity of the particle at the instant, then which one is correct?

Category: QuestionsThe distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax^2 + bx + c (a, b, c are positive constants). If v be the velocity of the particle at the instant, then which one is correct?
Editor">Editor Staff asked 11 months ago

The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax^2 + bx + c (a, b, c are positive constants). If v be the velocity of the particle at the instant, then which one is correct?
 
(a) Moves with retardation 2av^2
 
(b) Moves with retardation 2av^3
 
(c) Moves with acceleration 2av^3
 
(d) Moves with acceleration 2av^2
 
The question was asked in a national level competition.
 
My question comes from Calculus Application topic in division Application of Calculus of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

Correct option is (b) Moves with retardation 2av^3
 
Easy explanation: We have, t = ax^2 + bx + c  ……….(1)
 
Differentiating both sides of (1) with respect to x we get,
 
dt/dx = d(ax^2 + bx + c)/dx = 2ax + b
 
Thus, v = velocity of the particle at time t
 
= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)^-1  ……….(2)
 
Thus, acceleration of the particle at time t is,
 
= dv/dt = d((2ax + b)^-1)/dt
 
= -1/(2ax + b)^2 * 2av
 
= -v^2*2av   [as, v = 1/(2ax + b)]
 
= -2av^3
 
That is the particle is moving with retardation 2av^3.