A constant power of 7 watt is applied on a toy car of mass 15 kg. The distance (in m) travelled by the car when its velocity increases from 3ms^-1 to 5 ms^-1 is $\_\_\_\_\_\_$m . (answer to the nearest integer )

Given,

$$\begin{gathered} Power P=7W \\ mass m=15kg \\ {v}_i=3m/s \\ {v}_f=5m/s \end{gathered}$$

By the work energy theorem work done is

$$\begin{gathered} Work done=\frac{1}{2}m({v_f}^2-{v_i}^2) \\ W=\frac{1}{2}\times 15(25-9) \\ W=120J \end{gathered}$$

Time of work done,

$$\begin{gathered} t=\frac{work done}{power} \\ t=\frac{120}{7} \\ t=17.14s \\ Acceleration, \\ a=\frac{v_f-v_i}{t} \\ a=\frac{5-3}{17.14} \\ a=0.116m/s^2 \\ Dis\tan ce=\frac{{v_f}^2-{v_i}^2}{2a} \\ Dis\tan ce=\frac{5^2-3^2}{2\times 0.116} \\ Dis\tan ce=68.96 \\ Dis\tan ce=69m \end{gathered}$$