Irodov 1.4: Reading a Motion Graph's Three Speeds – Amit & Deepali Sinha Foundation

Problem Irodov 1.4

A point moves along a straight line, always in the same sense, so the distance it has covered, $s$ , increases monotonically with the elapsed time $t$ and never decreases. A plot of $s$ (meters, marked at $1.0$ and $2.0$ ) against $t$ (seconds, marked at $10$ and $20$ ) starts at the origin, stays close to zero for the first several seconds, rises steeply through an S-shaped bend, and then levels off toward a constant distance as $t$ grows further. Using such a plot, find:

the average velocity over the whole motion, from launch until the point comes to rest
the largest instantaneous velocity reached at any point of the motion
the time $t_0$ at which the instantaneous velocity equals the mean velocity averaged over the interval from launch up to $t_0$

The plot supplied with the problem: distance s in meters against time t in seconds. The curve leaves the origin, stays nearly flat, rises through an S-shaped bend, and levels off by the right edge. — The plot supplied with the problem: distance $s$ in meters against time $t$ in seconds. The curve leaves the origin, stays nearly flat, rises through an S-shaped bend, and levels off by the right edge.

Given

the plotted curve $s(t)$ (marked in m and s)

Find

(a)

\langle v\rangle

over the whole motion, (b)

v_{\mathrm{max}}

, (c)

t_0

Solution

Step 1Average velocity is the slope of the total chord

Over the whole motion, from launch at $(0,0)$ to the instant the curve becomes level (the point moves no further), the average velocity is the total distance divided by the total elapsed time, exactly the slope of the straight chord joining those two points on the plot.

\langle v\rangle = \frac{s_{\text{total}}}{t_{\text{total}}}

Step 2Maximum velocity is the slope of the steepest tangent

The instantaneous velocity is $v(t)=ds/dt$ , the slope of the tangent to the curve at $t$ ; it is largest exactly where the curve is steepest, at the inflection point (marked $P$ in the construction figure below) where the curve bends from concave up to concave down and $d^2s/dt^2=0$ .

v_{\mathrm{max}} = \left.\frac{ds}{dt}\right|_{P}

Step 3The crossover time is where a line from the origin is tangent to the curve

Write the running mean velocity as $\langle v\rangle(t)=s(t)/t$ and differentiate it by the quotient rule:

\frac{d}{dt}\left[\frac{s(t)}{t}\right] = \frac{t\,s'(t)-s(t)}{t^2}.

A fraction vanishes exactly where its numerator does, so setting this derivative to zero at $t=t_0$ leaves

t_0\,s'(t_0)-s(t_0) = 0 .

Dividing through by $t_0$ and writing $s'(t_0)=v(t_0)$ turns the stationary condition into part (c)’s requirement itself:

v(t_0) = \dfrac{s(t_0)}{t_0}

(1)

Geometrically, the left side is the curve’s own slope at $t_0$ and the right side is the slope of the chord from $O$ , so the two coincide only where a line through the origin touches the curve. That tangency point is marked $Q$ below.

Step 4Read the three constructions off the plot

The three constructions the steps justify: the whole-motion chord from O, the tangent at the steepest point P, and the line from O touching at Q, the tangency of . — The three constructions the steps justify: the whole-motion chord from $O$ , the tangent at the steepest point $P$ , and the line from $O$ touching at $Q$ , the tangency of (1).

The three quantities above are each a slope (or, for $t_0$ , a location) on the same drawing. Reading the chord’s rise and run against the marked scales gives the whole-motion average directly.

\langle v\rangle = \frac{2.0\text{ m}}{20\text{ s}}

\langle v\rangle = 10\text{ cm/s}

(2)

The tangent drawn through $P$ rises by the same amount, $0.25$ m, for every $1.0$ s of run along it, giving the maximum velocity the same way.

v_{\mathrm{max}} = \frac{0.25\text{ m}}{1.0\text{ s}}

v_{\mathrm{max}} = 25\text{ cm/s}

(3)

Unlike the first two, $t_0$ is not a slope but a location: it is simply $Q$ ’s own time coordinate, read directly off the horizontal scale where the origin-tangent construction of (1) touches the curve.

t_0 = 16\text{ s}

(4)

Check

The reading $t_0=16$ s falls after $P$ (the steepest point, near $12$ s), consistent with Step 3: the running mean $\langle v\rangle(t)$ climbs from near zero, peaks exactly at $t_0$ where it equals the local slope, then relaxes toward the eventual whole-motion average as $t$ grows. So the instantaneous velocity at $t_0$ must sit strictly between the whole-motion average and the peak, $10\text{ cm/s} < v(t_0) < 25\text{ cm/s}$ , which the reading is consistent with since $t_0$ lies between $P$ and where the curve levels off.

Remark

Step 3’s construction, a stationary point of $s(t)/t$ , is the same idea as the rule that average cost is minimized (or here, maximized) exactly where it equals marginal cost: any ratio of an accumulated quantity to elapsed time is stationary precisely where the instantaneous rate matches the running average, because that is the only instant a further increment neither pulls the average up nor down.