Kinematics

JEE Main level, single correct, medium

138 of 308

Question 138 of 308

The relation between time tt and distance x\mathrm{x} is t=ax2+bx\mathrm{t}=a \mathrm{x}^2+\mathrm{bx} where a\mathrm{a} and b\mathrm{b} are constants. The acceleration is

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Answer: (C) −2av3-2 a v^3

t=ax2+bxt=a x^2+b x
by differentiating acceleration
=−2av3=-2 a v^3

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