Irodov 1.21: Reversal Splits Distance from Displacement

Problem Irodov 1.21

A particle leaves the origin at $t=0$ moving in the positive $x$ direction, with velocity projection $v_x(t) = v_0(1-t/\tau)$ , where $v_0=10.0$ cm/s is the initial speed and $\tau=5.0$ s. Find:

the coordinate $x$ at $t=6.0$ , $10$ , and $20$ s
the moment(s) of time at which the particle is a distance $10.0$ cm from the origin
the total distance $s$ covered during the first $4.0$ s and during the first $8.0$ s, and a sketch of $s(t)$

Given

$v_0=10.0$ cm/s $=0.100$ m/s
$\tau=5.0$ s
$x(0)=0$

Find

(a)

x(6.0\text{ s})

x(10\text{ s})

x(20\text{ s})

; (b)

t

where

|x|=10.0

cm; (c)

s(4.0\text{ s})

s(8.0\text{ s})

Solution

Step 1Integrate the velocity to get the coordinate

The coordinate is the integral of $v_x(t)=v_0(1-t/\tau)$ from the start, $x=0$ at $t=0$ , to a general time $t$ .

x(t) = \int_0^t v_0\Bigl(1-\frac{t'}{\tau}\Bigr)dt' = v_0\left[t'-\frac{t'^2}{2\tau}\right]_0^t

x(t) = v_0\,t\left(1-\dfrac{t}{2\tau}\right)

(1)

Evaluating (1) at the three requested instants, with $v_0=0.100$ m/s and $2\tau=10$ s,

\begin{align*}x(6.0\ \text{s}) &= 0.100(6.0)\left(1-\frac{6.0}{10}\right)\\ x(10\ \text{s}) &= 0.100(10)\left(1-\frac{10}{10}\right)\\ x(20\ \text{s}) &= 0.100(20)\left(1-\frac{20}{10}\right)\end{align*}

\begin{gathered}x(6.0\ \text{s})=0.24\ \text{m}\\ x(10\ \text{s})=0\\ x(20\ \text{s})=-2.0\ \text{m}\end{gathered}

(2)

The coordinate x(t) of : it rises while v_x>0, peaks at t=τ=5 s, crosses back through the origin at t=2τ=10 s, and continues into negative x. The three marked ordinates are the part (a) values. — The coordinate $x(t)$ of (1): it rises while $v_x>0$ , peaks at $t=\tau=5$ s, crosses back through the origin at $t=2\tau=10$ s, and continues into negative $x$ . The three marked ordinates are the part (a) values.

Step 2Solve for the times at

|x|=10.0

Setting (1) equal to $+0.10$ m and to $-0.10$ m, and writing $2\tau=10$ s, gives two quadratics in $t$ .

\begin{align*}0.100\,t\left(1-\frac{t}{10}\right) &= 0.10 & 0.100\,t\left(1-\frac{t}{10}\right) &= -0.10\\ t^2-10t+10 &= 0 & t^2-10t-10&=0\\ t &= 5\pm\sqrt{15} & t&=5\pm\sqrt{35}\end{align*}

The first equation’s two roots are both positive; the second’s negative root is discarded, since the particle has not yet moved at $t<0$ .

\begin{aligned}t &= 5-\sqrt{15},\ 5+\sqrt{15},\ 5+\sqrt{35}\ \text{s}\\ &\approx 1.1,\ 8.9,\ 10.9\ \text{s}\end{aligned}

(3)

The distance covered s(t): it rises with x(t) up to the turnaround at t=τ, then keeps climbing past it. At τ the curve flattens to a momentary shoulder, not a kink, and the reversal shows only as a change of curvature. — The distance covered $s(t)$ : it rises with $x(t)$ up to the turnaround at $t=\tau$ , then keeps climbing past it. At $\tau$ the curve flattens to a momentary shoulder, not a kink, and the reversal shows only as a change of curvature.

Step 3Distance covered up to

t=4.0

s, before the turnaround

Since $\tau=5.0$ s, the particle is still moving in the positive direction throughout $0\le t\le4.0$ s, so the distance covered equals the coordinate itself, given by (1).

s(4.0\ \text{s}) = x(4.0\ \text{s}) = 0.100(4.0)\left(1-\frac{4.0}{10}\right)

s(4.0\ \text{s}) = 0.24\ \text{m}

(4)

Step 4Distance covered up to

t=8.0

s, past the turnaround

By (1), the coordinate reached at the turnaround is $x(\tau)=v_0\tau/2$ , which is also the distance covered so far, since the particle has moved only in the positive direction up to $\tau$ . Adding the speed integrated over the remaining stretch from $\tau$ to $t$ gives $s(t)$ beyond the turnaround.

\begin{aligned} s(t) &= x(\tau) + \int_\tau^t \Bigl[-v_0\Bigl(1-\frac{t'}{\tau}\Bigr)\Bigr]dt' \\ &= \frac{v_0\tau}{2} + v_0\left[\frac{t'^2}{2\tau}-t'\right]_\tau^t \\ &= \frac{v_0\tau}{2}\left[1+\left(1-\frac{t}{\tau}\right)^2\right] \end{aligned} \tag{5}

Evaluating at $t=8.0$ s, with $v_0\tau/2=0.25$ m,

s(8.0\ \text{s}) = 0.25\left[1+\left(1-\frac{8.0}{5.0}\right)^2\right]

s(8.0\ \text{s}) = 0.34\ \text{m}

(6)

Check

The formula (5) reduces to $v_0\tau/2$ at $t=\tau$ (the bracket becomes $1+0$ ), matching the $x(\tau)=v_0\tau/2$ of (1) exactly, so $s(t)$ is continuous through the turnaround as it must be. Also $s(8.0\text{ s})=0.34$ m exceeds $x(8.0\text{ s})=0.16$ m, exactly as it should once the particle has doubled back and $s$ counts the return leg that $x$ partly cancels.

Remark

The particle’s motion is uniformly decelerated ( $w_x=-v_0/\tau=\text{const}$ ), so $x(t)$ is one parabola throughout, but $s(t)$ is glued together from two different pieces at $t=\tau$ : before the turnaround it rides along with $x(t)$ , and after, it is built from $|v_x|$ instead of $v_x$ . The same split drove problem 1.20’s return-trip distance, and it is the generic shape of any $s(t)$ for motion that reverses once.