Question

A vehicle of mass 600 kg with an engine operating at constant power \(P\) accelerates from rest on a straight horizontal road. The vehicle covers a distance of 600 m in 1 minute. Neglecting all losses, the magnitude of \(P\) is ............. kW. (Round off to 2 decimal places)

Hint:

For constant power motion, velocity increases as \(v \propto t^{1/2}\) and distance as \(s \propto t^{3/2}\).

Answer 1

Correct Answer: 1.11

Step 1: Relation between power and velocity.
Since \(P = Fv = m a v\) and \(a = \frac{dv}{dt}\): \[ P = m v \frac{dv}{dt} \] \[ $\Rightarrow$ v \, dv = \frac{P}{m} dt \] Integrate both sides from \(0\) to \(v\) and \(0\) to \(t\): \[ \frac{v^2}{2} = \frac{P t}{m} $\Rightarrow$ v = \sqrt{\frac{2Pt}{m}} \]

Step 2: Relate distance and time.
\[ s = \int_0^t v\, dt = \int_0^t \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \frac{2}{3} t^{3/2} \] Rearrange for \(P\): \[ P = \frac{9m s^2}{8t^3} \]

Step 3: Substitute values.
\(m = 600\ \text{kg},\ s = 600\ \text{m},\ t = 60\ \text{s}\) \[ P = \frac{9 \times 600 \times 600^2}{8 \times 60^3} = \frac{9 \times 600 \times 360000}{8 \times 216000} = \frac{1.944 \times 10^9}{1.728 \times 10^6} = 1125\, \text{W} = 1.125\, \text{kW} \] Rounded: \(1.13\, \text{kW}\).

Step 4: Conclusion.
Hence, the power \(P = 1.13\, \text{kW}\).