Irodov 1.119: Square-Root Speed Hides Constant Acceleration

Problem Irodov 1.119

A locomotive of mass $m$ starts from rest and moves so that its speed depends on the distance $s$ it has covered as $v = a\sqrt{s}$ , with $a$ a given constant. Find the total work done by all the forces acting on the locomotive together during the first $t$ seconds of its motion.

The relation v=asqrt(s) between speed and distance covered, drawn schematically since a is not assigned a numerical value here. The marked point is the state reached after the elapsed time t used in the solution, with abscissa s(t) and ordinate v(t). — The relation $v=a\sqrt{s}$ between speed and distance covered, drawn schematically since $a$ is not assigned a numerical value here. The marked point is the state reached after the elapsed time $t$ used in the solution, with abscissa $s(t)$ and ordinate $v(t)$ .

Given

$v=a\sqrt{s}$
$m$

Find

A

during the first

t

seconds

Solution

Step 1Distance traveled from the given speed law

Since $v=ds/dt$ , the relation $v=a\sqrt s$ is a separable differential equation for $s(t)$ . Separating variables and integrating from the start of motion, where $s=0$ at $t=0$ , gives $s$ as an explicit function of time.

\begin{align*}\int_0^{s}\frac{ds'}{a\sqrt{s'}} &= \int_0^t dt' \\ \frac{2\sqrt{s}}{a} &= t\end{align*}

s(t) = \dfrac{a^2t^2}{4}

(1)

Step 2Speed reached after time

t

Substituting (1) back into the given law $v=a\sqrt{s}$ gives the speed at time $t$ .

v(t) = a\sqrt{s(t)} = a\sqrt{\dfrac{a^2t^2}{4}}

v(t) = \dfrac{a^2t}{2}

(2)

Step 3Work from the work-energy theorem

The locomotive starts from rest, so the total work done by every force acting on it equals the kinetic energy it has gained by time $t$ ; substituting (2) gives the algebraic form.

\begin{align*}A &= \frac12 mv(t)^2-0 \\ &= \frac12 m\left(\frac{a^2t}{2}\right)^2\end{align*}

A = \dfrac{ma^4t^2}{8}

(3)

Check

Since $[a]=[v]/\sqrt{[s]}=\mathrm{m^{1/2}/s}$ , the boxed combination carries $[m][a]^4[t]^2=\mathrm{kg}\cdot(\mathrm{m^{1/2}/s})^4\cdot\mathrm{s^2}=\mathrm{kg\,m^2/s^2}$ , an energy, as required.

Remark

Differentiating (2) gives $dv/dt=a^2/2$ , a constant: although the given law $v=a\sqrt s$ looks like a nontrivial relation, it describes plain uniformly accelerated motion from rest with acceleration $w=a^2/2$ . That is why the boxed work matches the familiar form $mw^2t^2/2$ for a body started from rest under constant acceleration $w$ .