Question

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

Answer 1

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.