A body moves in the positive x -direction so that its speed…

A body moves in the positive $x$-direction so that its speed depends on its position as $v(x) = b\sqrt x$, where $b$ is a positive constant. Find: (a) its velocity and acceleration as functions of time $t$; (b) its average speed as a function of the distance travelled $x$.

Explanation of Method

Speed that depends on position: v = b√x.

Step 1: Position as a function of time

Velocity is $v = \dfrac{dx}{dt}$, so $\dfrac{dx}{dt} = b\sqrt x$. Separate the variables and integrate, taking $x = 0$ at $t = 0$: \[\int_0^x \frac{dx}{\sqrt x} = \int_0^t b\,dt \;\Longrightarrow\; 2\sqrt x = bt \;\Longrightarrow\; x = \frac{b^2 t^2}{4}.\]

Step 2: (a) Velocity and acceleration

\[ v(t) = b\sqrt{x} = b \cdot \frac{bt}{2} = \frac{b^2}{2}\,t, \]

\[ a = \frac{dv}{dt} = \frac{b^2}{2}. \]

\[v(t) = \frac{b^2}{2}\,t, \qquad a = \frac{b^2}{2} \ \text{(constant)}\]

Step 3: (b) Average speed over the distance $x$

From step 1, the time to reach $x$ is $t = \dfrac{2\sqrt x}{b}$, so

\[ \bar v = \frac{x}{t} = \frac{x}{2\sqrt x / b}. \]

\[\bar v(x) = \frac{b}{2}\sqrt x\]

Check without calculus

Squaring $ v = b\sqrt x $ gives $ v^2 = b^2 x = 2 \cdot \tfrac{b^2}{2} \cdot x $, which is $ v^2 = 2ax $ with $ a = \tfrac{b^2}{2}\ \checkmark $

For constant acceleration from rest, the average speed is half the final speed: $ \bar v = \tfrac12 b\sqrt x\ \checkmark $

$x$ $v$ $v = b\sqrt x$ $\bar v = \tfrac{b}{2}\sqrt x$ $\text{speed vs position}$ $t$ $v$ $v = \tfrac{b^2}{2}t$ $\bar v = \tfrac{v}{2}$ $\text{speed vs time}$ $\text{area} = x$

speed
average speed

Left: speed grows like $\sqrt x$ with position. Right: the same motion against time is a straight line, i.e. constant acceleration from rest, so the average speed is always half the current speed. (Drawn for $b = 2$.)

Result

\[\boxed{\,v(t) = \frac{b^2}{2}\,t, \qquad a = \frac{b^2}{2} \ \text{(constant)}\,}\]

\[\boxed{\,\bar v(x) = \frac{b}{2}\sqrt x\,}\]

Remark

Note. Although the law $v = b\sqrt x$ looks unusual, it describes the most ordinary motion there is: uniform acceleration from rest. The formula $v^2 = 2ax$ is the key to recognising it.