A car of mass M accelerates starting from rest. Velocity of the car is given by $V = {(\frac{2 Pt}{M})}^{1 / 2}$ where ‘P’ is the constant power supplied by engine. The position of a car as a function of time is given by

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${(\frac{8 P}{9 M})}^{1 / 2} t^{3 / 2}$

${(\frac{9 P}{8 M})}^{1 / 2} t^{3 / 2}$

${(\frac{8 P}{9 M})}^{1 / 2} t^{2 / 3}$

$(\frac{9 P}{8 M}) t^{3}$

answer is A.

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Detailed Solution

$v = {(\frac{2 Pt}{m})}^{1 / 2} \Rightarrow \frac{ds}{dt} = {(\frac{2 Pt}{m})}^{1 / 2} \Rightarrow ds = {(\frac{2 P}{m})}^{1 / 2} t^{1 / 2} \cdot dt On integrating s = {(\frac{8 P}{9 m})}^{\frac{1}{2}} t^{\frac{3}{2}}$

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