A point moves rectilinearly so that its velocity is
A point moves rectilinearly so that its velocity is proportional to the square root of the distance travelled, $v = c\sqrt{s}$. Show that the motion occurs under a force of constant magnitude.
Explanation of Method
Differentiating with respect to distance instead of time.
Step 1: The acceleration in terms of distance
The speed is given as a function of $s$, not of $t$, so the chain rule of Problem 329 is the natural tool: \[w = \frac{dv}{dt} = \frac{dv}{ds}\cdot\frac{ds}{dt} = v\,\frac{dv}{ds}. \tag{1}\]
Step 2: Differentiate the given law
\[ v = c\,s^{1/2} \;\Longrightarrow\; \frac{dv}{ds} = \frac{c}{2\sqrt{s}}. \]
Step 3: Combine
The two square roots cancel exactly: $w = c\sqrt{s}\cdot\frac{c}{2\sqrt{s}} = \frac{c^{2}}{2}$.
The result contains neither $s$ nor $t$, so the acceleration is constant and, by Newton's second law, so is the force $F = mw = \dfrac{mc^{2}}{2}$.
Step 4: Verification
\[\text{Uniform acceleration from rest: } s = \frac{wt^{2}}{2},\ v = wt \;\Longrightarrow\; v = w\sqrt{\frac{2s}{w}} = \sqrt{2ws}\]
\[\text{So } c = \sqrt{2w}, \text{ i.e. } w = \frac{c^{2}}{2} \;\;\checkmark \ \text{— the converse statement}\]
\[\text{Explicitly, } \frac{ds}{dt} = c\sqrt{s} \text{ integrates to } s = \frac{c^{2}t^{2}}{4}, \text{ whence } \ddot s = \frac{c^{2}}{2} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,w = \dfrac{c^{2}}{2} = \text{const} \;\Longrightarrow\; F = \dfrac{mc^{2}}{2} = \text{const}\,}\]
Remark
Remark. The law $v = \sqrt{2ws}$ is the familiar $v^{2} = 2ws$ for motion from rest under constant acceleration, so the problem is that relation read backwards. The same computation shows the general pattern: $v \propto s^{n}$ gives $w \propto s^{2n-1}$, constant precisely when $n = \tfrac12$.