A constant power of \(7 \mathrm{~W}\) is supplied on a toy car of mass \(15 \mathrm{~kg}\). The distance…
A constant power of \(7 \mathrm{~W}\) is supplied on a toy car of mass \(15 \mathrm{~kg}\). The distance travelled by the car when its velocity increases from \(3 \mathrm{~ms}^{-1}\) to \(5 \mathrm{~ms}^{-1}\) is
\(56 \mathrm{~m}\)
\(7 \mathrm{~m}\)
\(61 \mathrm{~m}\)
\(70 \mathrm{~m}\)
Solution
Given, \(P=7 \mathrm{~W}\), mass, \(m=15 \mathrm{~kg}\),
\(v_i=3 \mathrm{~ms}^{-1} \text { and } v_f=5 \mathrm{~ms}^{-1}\)
From work-energy theorem,
\(\begin{array}{rlrl}
& \text { Work done } & =\frac{1}{2} m\left(v_f^2-v_i^2\right) \\
& \Rightarrow & & =\frac{1}{2} \times 15(25-9) \\
& \Rightarrow & W & =120 \mathrm{~J}
\end{array}\)
So, the time of work done,
\(t=\frac{\text { work }}{\text { power }}=\frac{120}{7}=1714 \mathrm{~s}\)
Hence, the acceleration,
\(a=\frac{v_f-v_i}{t}=\frac{5-3}{17.14}=0.116 \mathrm{~m} / \mathrm{s}^2\)
Distance travelled by car,
\(D=\frac{v_f^2-v_i^2}{2 a}=\frac{5^2-3^2}{2 \times 0.116}=68.96 \approx 70 \mathrm{~m}\)
Hence, the correct option is (d).