Essential University Physics: Volume 1 (4th Edition) Clone

a) $v = \sqrt{\frac{2Pt}{M}}$ b) $x = \frac{2}{3}\sqrt{\frac{2Pt}{M}}t^{3/2}$
a) We know that work is equal to the power times the change in time. Since all work goes to kinetic energy for the train, it follows: $\frac{1}{2}Mv^2 = Pt \\ v = \sqrt{\frac{2Pt}{M}}$ (This assumes the train starts at rest. If it did not, a $v_0^2$ term would have to be added under the square root.) Calling the initial position x=0, we take the integral to find: $x = \int_0^x \sqrt{\frac{2Pt}{M}}$ $x = \frac{2}{3}\sqrt{\frac{2Pt}{M}}t^{3/2}$