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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1 Answers
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