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A particle is moving along the path given by $y = {Ct}^{6}$ (where C is a positive constant). The relation between the acceleration ‘a’ and the velocity (v) of the particle at t=5 is

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$5 a = v$

$v = 5 a$

$a = \sqrt{v}$

$a = v$

answer is D.

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Detailed Solution

$\begin{array}{rcl} velocity v & = & 6 c t^{5} - - - (1) \\ acceleration a & = & 30 {Ct}^{4} - - - - - (2) \\ dividing we get \frac{v}{a} & = & \frac{t}{5} \\ if t & = & 5, v = a \end{array}$

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