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
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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)]
That is the particle is moving with retardation 2av^3.