Irodov 1.18: One Graph Breeds Three More – Amit & Deepali Sinha Foundation

Problem Irodov 1.18

A point moves along the $x$ axis. Its velocity projection $v_x$ , as a function of time $t$ in seconds, is piecewise linear through the corners $(0,0)$ , $(1,1)$ , $(3,1)$ , $(6,-2)$ , $(7,0)$ : it starts from rest and climbs straight to $v_x=1$ at $t=1$ s, holds that value to $t=3$ s, then falls in a straight line through zero at $t=4$ s down to $v_x=-2$ at $t=6$ s, and finally climbs straight back to $v_x=0$ at $t=7$ s. The point starts at $x=0$ at $t=0$ . Sketch, as functions of $t$ over $0\le t\le7$ s: the acceleration projection $w_x$ , the coordinate $x$ , and the total distance covered $s$ .

Given

$v_x(t)$ piecewise linear as plotted
$x(0)=0$

Find

sketches of

w_x(t)

x(t)

s(t)

Solution

Step 1Read the velocity law off the graph

Each straight run of the given plot is a linear function of $t$ , fixed by the two corners it joins; writing down all four gives $v_x(t)$ as a formula rather than only a picture.

v_x(t) = \begin{cases} t, & 0\le t\le 1\\ 1, & 1\le t\le 3\\ 4-t, & 3\le t\le 6\\ 2t-14, & 6\le t\le 7 \end{cases} \tag{1}

The first run is the one that is easy to overlook: the point leaves the origin at rest and is brought up to unit speed over the first second, so the motion opens with an acceleration rather than with a velocity already established.

The three derived sketches, left to right: acceleration w_x(t), a four level step; coordinate x(t), flat at the origin, peaking at t=4, back to zero at t=7; and distance covered s(t), climbing throughout with a shoulder at t=4. — The three derived sketches, left to right: acceleration $w_x(t)$ , a four level step; coordinate $x(t)$ , flat at the origin, peaking at $t=4$ , back to zero at $t=7$ ; and distance covered $s(t)$ , climbing throughout with a shoulder at $t=4$ .

Step 2Acceleration is the slope of each run

Differentiating each piece of (1) in turn,

w_x(t) = \frac{dv_x}{dt} = \begin{cases} 1, & 0<t<1\\ 0, & 1<t<3\\ -1, & 3<t<6\\ 2, & 6<t<7 \end{cases}

A four level step:

+1

, then

0

, then

-1

, then

+2

the first run starting the point off, the third slowing it down and reversing it, the last one braking that reversal, sketched in the left panel of the figure above.

Step 3Coordinate from integrating

v_x

piece by piece

Starting from $x(0)=0$ and carrying the running total across each corner keeps $x(t)$ continuous even though $v_x(t)$ has corners in it.

\begin{aligned} x(t) &= \int_0^t \tau\,d\tau = \frac{t^2}{2}, & 0\le t\le1 \\[4pt] x(t) &= x(1)+\int_1^t d\tau = 0.5+(t-1) = t-0.5, & 1\le t\le3 \\[4pt] x(t) &= x(3)+\int_3^t (4-\tau)\,d\tau = 2.5+\Bigl[4\tau-\dfrac{\tau^2}{2}\Bigr]_3^t = 4t-\frac{t^2}{2}-5, & 3\le t\le6 \\[4pt] x(t) &= x(6)+\int_6^t (2\tau-14)\,d\tau = 1+\Bigl[\tau^2-14\tau\Bigr]_6^t = (t-7)^2, & 6\le t\le7 \end{aligned} \tag{2}

Evaluating at the corners gives the values the sketch has to pass through.

x(1)=0.5, \qquad x(3)=2.5, \qquad x(4)=3\ (\text{maximum, where } v_x=0), \qquad x(6)=1, \qquad x(7)=0

Rises to

3

t=4

, falls back to

0

t=7

The sketch is the middle panel of the figure above, and it leaves the origin flat, not at a slope, because $v_x(0)=0$ .

Step 4Distance covered from the running total of

\lvert v_x\rvert

Up to $t=4$ the point moves only in the positive direction, so $s(t)=x(t)$ by (2); beyond $t=4$ each further step adds $\lvert v_x\rvert\,dt$ to the running total.

\begin{align*}s(t) &= s(4)+\int_4^t (\tau-4)\,d\tau = 3+\frac{(t-4)^2}{2}, & 4\le t\le6 \\[4pt] s(t) &= s(6)+\int_6^t (14-2\tau)\,d\tau = 5+\Bigl[14\tau-\tau^2\Bigr]_6^t = 14t-t^2-43, & 6\le t\le7\end{align*}

s(1)=0.5, \qquad s(3)=2.5, \qquad s(4)=3, \qquad s(6)=5, \qquad s(7)=6

Climbs throughout, reaching

6

t=7

The sketch is the right panel of the figure above.

Check

The areas under the given plot settle both endpoints independently of the integrations. Above the axis, from $t=0$ to $t=4$ , the trapezium of the first three runs has area $\frac{1}{2}(1)(1)+(2)(1)+\frac{1}{2}(1)(1)=3$ ; below it, from $t=4$ to $t=7$ , the two triangles give $\frac{1}{2}(2)(2)+\frac{1}{2}(1)(2)=3$ . The two are equal, so the signed total is $x(7)=0$ and the unsigned total is $s(7)=3+3=6$ , exactly the endpoints the two right hand sketches carry. Up to $t=4$ the curves coincide, $s=x$ ; past it $s>x$ , as it must once the direction reverses.

Remark

Of the three sketches, only $x(t)$ keeps $v_x(t)$ in full: differentiating it recovers $v_x$ exactly, sign included. $w_x=dv_x/dt$ is a derivative, not an antiderivative, and being a step it forgets the level $v_x$ stepped from, keeping only the slope, which is why the same step would serve a point that started at any speed. And $s(t)$ , built from $\lvert v_x\rvert$ rather than $v_x$ , forgets the sign entirely: the reversal at $t=4$ leaves no trace in $s(t)$ alone, since $ds/dt=\lvert v_x\rvert$ never goes negative, only in $x(t)$ , which brings the point home to where it started.