Irodov 1.20: Turning Point Doubles the Distance – Amit & Deepali Sinha Foundation

Problem Irodov 1.20

A particle’s position vector is $\vect r(t) = \vect a\, t(1-\alpha t)$ , where $\vect a$ is a fixed constant vector and $\alpha$ is a positive constant, and the particle is at the origin at $t=0$ . Find:

the velocity $\vect v(t)$ and the acceleration $\vect w(t)$
the time interval $\Delta t$ after which the particle returns to the origin, and the total distance $s$ it covers over that interval

Given

$\vect r(t) = \vect a\,t(1-\alpha t)$
$\alpha>0$ constant

Find

(a)

\vect v(t)

\vect w(t)

; (b)

\Delta t

s

Solution

Step 1Velocity and acceleration by differentiation

Since $\vect a$ and $\alpha$ are constants, differentiating the given $\vect r(t)$ term by term gives the velocity, and differentiating again gives the acceleration.

\vect v(t) = \frac{d\vect r}{dt} = \frac{d}{dt}\bigl[\vect a\,t-\vect a\,\alpha t^2\bigr]

\vect v(t) = \vect a\,(1-2\alpha t)

(1)

\vect w(t) = \frac{d\vect v}{dt}

\vect w(t) = -2\alpha\vect a

(2)

Step 2The return time from the zero of

\vect r(t)

Besides the launch instant $t=0$ , the scalar factor $t(1-\alpha t)$ has one other zero.

t(1-\alpha t) = 0 \;\Longrightarrow\; t=0 \ \text{or}\ t=\frac{1}{\alpha}

\Delta t = \dfrac{1}{\alpha}

(3)

Step 3Total distance from the turning point

The velocity (1) vanishes at the turning time $t_1$ .

\vect a\,(1-2\alpha t_1) = \vect 0 \;\Longrightarrow\; t_1 = \frac{1}{2\alpha}

Evaluating $\vect r(t)$ at $t_1$ gives the displacement reached before the reversal, whose magnitude the distance is built from.

\begin{align*}\vect r(t_1) &= \vect a\, t_1(1-\alpha t_1) \\ &= \vect a\cdot\frac{1}{2\alpha}\left(1-\frac{1}{2}\right) \\ &= \frac{\vect a}{4\alpha}\end{align*}

The scalar component of r along a against time (shape only, not to scale). The peak sits at the turning time t_1=1/2α, and the curve returns to O at Δ t=1/α. — The scalar component of $\vect r$ along $\vect a$ against time (shape only, not to scale). The peak sits at the turning time $t_1=1/2\alpha$ , and the curve returns to $O$ at $\Delta t=1/\alpha$ .

The total distance is twice this magnitude, once outbound and once on the return.

s = 2\left|\vect r(t_1)\right| = 2\cdot\frac{a}{4\alpha}

s = \dfrac{a}{2\alpha}

(4)

Check

Dimensionally $[a]/[\alpha]$ is a length, since $\alpha$ carries units of inverse time and $\vect r=\vect a\,t(1-\alpha t)$ requires $\alpha t$ to be dimensionless, so $[\alpha]=1/[t]$ and $s=a/(2\alpha)$ is a length as required. Also $\vect r(\Delta t)=\vect a\,\Delta t(1-\alpha\Delta t)=\vect a\cdot\frac{1}{\alpha}\cdot(1-1)=\vect 0$ , confirming the particle is back at the origin exactly at $\Delta t$ .

Remark

The distance $s=a/(2\alpha)$ is twice the peak displacement $a/(4\alpha)$ , never the net displacement, which is zero at $\Delta t$ : a return trip is the standard trap where displacement and distance covered diverge, here made explicit because $\vect r(t)$ is a single quadratic-in- $t$ factor whose zero and whose turning point are two different, easily confused instants.