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 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^2The 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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Answer» 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. |
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