InterviewSolution
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InterviewSolution
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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 = -4atI 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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Answer» 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^2x0^2 – 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 |
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