# 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 and u(≠0) be the initial velocity of the particle 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 and u(≠0) be the initial velocity of the particle 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 and u(≠0) be the initial velocity of the particle then which one is correct?

(a) 1/v^2 + 1/u^2 = 4at

(b) 1/v^2 + 1/u^2 = -4at

(c) 1/v^2 – 1/u^2 = 4at

(d) 1/v^2 – 1/u^2 = -4at

I got this question in examination.

The above asked question is from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

Correct choice is (c) 1/v^2 – 1/u^2 = 4at

The explanation is: 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)

Initially, when t = 0 and v = u, let x = x0; hence, from (1) we get,

ax0^2 + bx0 + c = 0

Or ax0^2 + bx0 = -c   ……….(3)

And from (2) we get, u = 1/(2ax0 + b)

Thus, 1/v^2 – 1/u^2 = (2ax + b)^2 – (2ax0 + b)^2

= 4a^2x^2 + 4abx – 4a^2×0^2 – 4abx0

= 4a^2x^2 + 4abx – 4a(ax0^2 – bx0)

= 4a^2x^2 + 4abx – 4a(-c)   [using (3)]

= 4a(ax^2 + bx + c)

Or 1/v^2 – 1/u^2 = 4at