A tram of mass m moves so that its speed depends on the…

A tram of mass $m$ moves so that its speed depends on the distance travelled as $v = ks^{3/2}$ ($k$ constant). Find (a) its acceleration, (b) the work of all forces acting on it.

Explanation of Method

Work on a tram whose speed grows with distance.

Step 1: (a) Acceleration

\[ a = \frac{dv}{dt} = \frac{dv}{ds}\,v = \tfrac32 k s^{1/2} \cdot k s^{3/2}. \]

\[a = \tfrac32 k^2 s^2\]

Step 2: (b) Work

The work of all forces equals the change in kinetic energy: $W = \tfrac12 m v^2 = \tfrac12 m k^2 s^3$.

Step 3: In terms of time?

With this law, $ds/dt = ks^{3/2}$, and $\int_0 ds/s^{3/2}$ diverges: starting from rest at $s = 0$, the tram would need infinite time to move at all. So $W$ cannot be written as a function of the time from the start. In the standard version of this problem the law is $v = k\sqrt{s}$; then $a = k^2/2$ is constant, $v = k^2t/2$, and

\[ W(t) = \tfrac12 m v^2 = \frac{mk^4t^2}{8}. \]

$v,\ a$ $s$ $v = ks^{3/2}$ $a = \tfrac32 k^2 s^2$

Speed (blue) and acceleration (red, dashed) against distance, in arbitrary units. Both start from zero at $s = 0$, which is why the motion can never begin from rest with exactly this law.

Result

\[\boxed{\,a = \tfrac32 k^2 s^2\,}\]

\[\boxed{\,W = \tfrac12 m v^2 = \tfrac12 m k^2 s^3\,}\]

Remark

Check. The original answers (a) $\tfrac32 k^2 s^2$ and $W = \tfrac12 mk^2s^3$ are correct. The question asks for the work "during the time $t$", which the original left open; as shown above, that is impossible with $s^{3/2}$ starting from rest, and the likely intended law $v = k\sqrt s$ gives $W = mk^4t^2/8$.