Irodov 1.22: Square-Root Speed, Quadratic Position – Amit & Deepali Sinha Foundation

Problem Irodov 1.22

A particle moves along the positive $x$ axis with its speed set by its own position, $v=\alpha\sqrt{x}$ , where $\alpha$ is a positive constant. At $t=0$ the particle sits at $x=0$ . Find:

the velocity $v(t)$ and the acceleration $w(t)$ as functions of time
the mean velocity $\langle v\rangle$ , averaged over the time taken to cover the first $s$ meters of the path

Given

$v=\alpha\sqrt{x}$
$x(0)=0$ at $t=0$

Find

(a)

v(t)

w(t)

; (b)

\langle v\rangle

over the first

s

meters

Solution

Step 1Separate variables and integrate to find

x(t)

The velocity $dx/dt$ is given only as a function of $x$ , so the equation separates directly into $x$ and $t$ ; integrating both sides from the start, $x=0$ at $t=0$ , to a general instant fixes $x(t)$ .

\frac{dx}{dt} = \alpha\sqrt{x} \;\Longrightarrow\; \int_0^{x} \frac{dx'}{\sqrt{x'}} = \int_0^{t} \alpha\,dt'

\begin{align*}2\sqrt{x} &= \alpha t \\ x &= \frac{\alpha^2 t^2}{4}\end{align*}

x(t) = \dfrac{\alpha^2 t^2}{4}

(1)

Step 2Differentiate for the velocity and the acceleration

With $x(t)$ in hand (1), the velocity and acceleration follow by straightforward differentiation.

v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{\alpha^2 t^2}{4}\right)

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

(2)

w(t) = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{\alpha^2 t}{2}\right)

w(t) = \dfrac{\alpha^2}{2}

(3)

so the particle moves with constant acceleration even though its speed was only ever given as a function of position, never of time.

Step 3Mean velocity over the first

s

meters

Setting $x(t_s)=s$ in (1) and solving for $t_s$ ,

s = \frac{\alpha^2 t_s^2}{4} \;\Longrightarrow\; t_s = \frac{2\sqrt{s}}{\alpha}.

The mean velocity is the distance $s$ divided by this time,

\begin{align*}\langle v\rangle &= \frac{s}{t_s} \\ &= \frac{s}{2\sqrt{s}/\alpha}\end{align*}

\langle v\rangle = \dfrac{\alpha\sqrt{s}}{2}

(4)

the slope of the chord $OP$ drawn below.

The position x(t) of , a parabola through the origin. The point P is where the particle has covered the distance s, at t_s=2 s/α; the chord OP has slope s/t_s, the mean velocity. — The position $x(t)$ of (1), a parabola through the origin. The point $P$ is where the particle has covered the distance $s$ , at $t_s=2\sqrt s/\alpha$ ; the chord $OP$ has slope $s/t_s$ , the mean velocity.

Check

Substituting $x(t)$ from (1) back into $\alpha\sqrt{x}$ reproduces $v(t)$ exactly: $\alpha\sqrt{\alpha^2t^2/4} = \alpha\cdot(\alpha t/2) = \alpha^2 t/2$ , matching Step 2, since $t\geq0$ keeps the square root’s sign consistent. And $\langle v\rangle\to0$ as $s\to0$ , as it must for a particle starting from rest.

Remark

A speed proportional to $\sqrt{x}$ , starting from rest at $x=0$ , always integrates to uniformly accelerated motion: this is the same $x\propto t^2$ law a body dropped from rest obeys, with $w=\alpha^2/2$ standing in for $g$ . The unusual part of the problem, a velocity law given in $x$ rather than $t$ , was really just uniform acceleration wearing a disguise.