Irodov 1.23: A Deceleration That Actually Stops – Amit & Deepali Sinha Foundation

Problem Irodov 1.23

A point moves along a straight line, slowing down with a deceleration whose magnitude is $w = \alpha\sqrt{v}$ , where $v$ is the point’s instantaneous speed and $\alpha$ is a positive constant. At the initial moment its speed is $v_0$ . Find the distance the point covers before it comes to rest, and the time that takes.

Given

$v_0$
$\alpha$

Find

the stopping distance

s

, the stopping time

t

Solution

Step 1Write the deceleration as a rate of change of speed

The point loses speed, so its acceleration is the negative of the stated magnitude, giving a first order equation for $v(t)$ .

\frac{dv}{dt} = -\alpha\sqrt{v} \tag{1}

Step 2Distance from a

v\,dv/ds

substitution

Writing the acceleration instead as $v\,dv/ds$ trades time for the coordinate $s$ along the path, after which the two sides separate directly.

\begin{align*}v\frac{dv}{ds} &= -\alpha\sqrt{v} \\ \sqrt{v}\,dv &= -\alpha\,ds\end{align*}

Integrating from the start ( $s=0$ , $v=v_0$ ) to the stop ( $s=s$ , $v=0$ ) gives the stopping distance.

\begin{align*}\int_{v_0}^{0}\sqrt{v}\,dv &= -\alpha\int_0^{s}ds \\ \left.\frac{2}{3}v^{3/2}\right|_{v_0}^{0} &= -\alpha s \\ -\frac{2}{3}v_0^{3/2} &= -\alpha s\end{align*}

s = \dfrac{2}{3\alpha}v_0^{3/2}

(2)

Step 3Time from separating

v

and

t

directly

Returning to (1) and separating variables gives the elapsed time to reach speed zero.

\begin{align*}\frac{dv}{\sqrt{v}} &= -\alpha\,dt \\ \int_{v_0}^{0}v^{-1/2}\,dv &= -\alpha\int_0^{t}dt \\ \left.2\sqrt{v}\right|_{v_0}^{0} &= -\alpha t \\ -2\sqrt{v_0} &= -\alpha t\end{align*}

t = \dfrac{2\sqrt{v_0}}{\alpha}

(3)

The speed against time that integrates to, with the stopping instant marked t_s. The curve meets the axis tangentially, since the deceleration dies away with the speed itself. — The speed against time that (1) integrates to, with the stopping instant marked $t_s$ . The curve meets the axis tangentially, since the deceleration dies away with the speed itself.

Check

Dimensionally, $[\alpha]=[w]/\sqrt{[v]}=(\mathrm{m/s^2})/\sqrt{\mathrm{m/s}}=\mathrm{m}^{1/2}\mathrm{s}^{-3/2}$ , so $\sqrt{v_0}/\alpha$ carries $\sqrt{\mathrm{m/s}}\cdot\mathrm{m}^{-1/2}\mathrm{s}^{3/2}=\mathrm{s}$ , a time, and $v_0^{3/2}/\alpha$ carries $(\mathrm{m/s})^{3/2}\cdot\mathrm{m}^{-1/2}\mathrm{s}^{3/2}=\mathrm{m}$ , a length, as required.

Remark

Because $w\propto\sqrt{v}$ rather than $w\propto v$ , the point stops in a finite time: $\int_0^{v_0}dv/\sqrt{v}$ converges at $v=0$ , where the earlier figure’s curve meets the axis with zero slope. A deceleration proportional to $v$ itself gives the classic $v=v_0e^{-\alpha t}$ instead, whose integral $\int_0^{v_0}dv/v$ diverges at $v=0$ , so that point only ever approaches rest and never actually arrives.