A body slows down with an acceleration that depends on its…

A body slows down with an acceleration that depends on its speed as $a(v) = -\dfrac{k}{\sqrt v}$, where $k$ is a positive constant. Its initial speed is $v_0$. (a) How long does it take to stop? (b) What distance does it cover in that time?

Explanation of Method

Braking with acceleration that depends on speed.

Step 1: (a) Time to stop

Write $a = \dfrac{dv}{dt}$, separate the variables and integrate from $v_0$ (at $t = 0$) to $v$: $\sqrt v\,dv = -k\,dt \;\Longrightarrow\; \tfrac23\left(v^{3/2} - v_0^{3/2}\right) = -kt$.

So the speed decreases as

\[ v^{3/2} = v_0^{3/2} - \tfrac32 kt. \]

The body stops when $v = 0$: $t_s = \frac{2\,v_0^{3/2}}{3k}$.

$v$ $t$ $v_0$ $0$ $t_s = \tfrac{2v_0^{3/2}}{3k}$ $\text{area} = s = \tfrac{2v_0^{5/2}}{5k}$ $v = \left(v_0^{3/2} - \tfrac32 kt\right)^{2/3}$

actual v(t)
uniform braking, same stopping time

Speed–time graph (drawn for $v_0 = 1$, $k = 1$). The deceleration $k/\sqrt v$ is small at first and grows as the body slows, so the curve bends down more and more steeply. The area under it is the stopping distance; it is larger than for uniform braking over the same time (dashed line).

Step 2: (b) Stopping distance

Here it is easier to use position instead of time. By the chain rule $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$, so

\[ v\,\frac{dv}{dx} = -\frac{k}{\sqrt v} \;\Longrightarrow\; v^{3/2}\,dv = -k\,dx. \]

Integrate from $v_0$ down to $0$ while $x$ goes from $0$ to $s$: $\int_{v_0}^{0} v^{3/2}\,dv = -k\,s \;\Longrightarrow\; -\tfrac25 v_0^{5/2} = -k\,s$.

\[s = \frac{2\,v_0^{5/2}}{5k}\]

Check with units

From $a = k/\sqrt v$, the constant $k$ has units $\mathrm{m\,s^{-2}}\cdot(\mathrm{m\,s^{-1}})^{1/2} = \mathrm{m^{3/2}\,s^{-5/2}}$.

$ \dfrac{v_0^{3/2}}{k}: \ \dfrac{\mathrm{m^{3/2}\,s^{-3/2}}}{\mathrm{m^{3/2}\,s^{-5/2}}} = \mathrm{s}\ \checkmark $ (a time)

$ \dfrac{v_0^{5/2}}{k}: \ \dfrac{\mathrm{m^{5/2}\,s^{-5/2}}}{\mathrm{m^{3/2}\,s^{-5/2}}} = \mathrm{m}\ \checkmark $ (a length)

Result

\[\boxed{\,t_s = \frac{2\,v_0^{3/2}}{3k}\,}\]

\[\boxed{\,s = \frac{2\,v_0^{5/2}}{5k}\,}\]

Remark

About the original solution. Its method (integrating $v(t)$ over time) is correct and gives $s = \tfrac{2}{3k}\cdot\tfrac35\left(v_0^{3/2}\right)^{5/3}$. But $\left(v_0^{3/2}\right)^{5/3} = v_0^{\frac32\cdot\frac53} = v_0^{5/2}$, not $v_0^3$. So the correct distance is $\tfrac{2v_0^{5/2}}{5k}$; the final answer $\tfrac{2v_0^3}{5k}$ does not even have the units of length. Also, the problem calls the acceleration "uniformly variable", but it is not uniform: it depends on the speed and grows as the body slows down.