Irodov 1.2: Half Distance, Half Time, Different Means

Problem Irodov 1.2

A point moves along a straight line. It covers the first half of the total distance at a constant velocity $v_0$ . It then covers the second half of the distance in two equal time intervals: during the first interval it moves at a constant velocity $v_1$ , and during the second interval at a constant velocity $v_2$ . Find the mean velocity of the point, averaged over the entire time of motion.

Distance covered versus time, drawn for illustrative speeds. The first leg has slope v_0 and reaches s at t_0; the two later legs, of equal duration and slopes v_1 then v_2, carry the curve to 2s. — Distance covered versus time, drawn for illustrative speeds. The first leg has slope $v_0$ and reaches $s$ at $t_0$ ; the two later legs, of equal duration and slopes $v_1$ then $v_2$ , carry the curve to $2s$ .

Given

$v_0$
$v_1$
$v_2$

Find

The mean velocity

\langle v\rangle

over the whole trip

Solution

Step 1Time the first half of the distance

Let the total distance be $2s$ . The first half, of length $s$ , is covered entirely at $v_0$ , so it takes

t_0 = \frac{s}{v_0}. \tag{1}

Step 2Time the second half of the distance

Call the time spent on the second half $t_1$ . Half of that time, $t_1/2$ , is spent at $v_1$ and the other half at $v_2$ , so the distance covered in the second leg is

v_1\cdot\frac{t_1}{2} + v_2\cdot\frac{t_1}{2} = s. \tag{2}

Solving for $t_1$ ,

t_1 = \dfrac{2s}{v_1+v_2}

(3)

Step 3Assemble the mean velocity

The mean velocity over the whole trip is total distance over total time,

\langle v\rangle = \frac{2s}{t_0+t_1}. \tag{4}

Substituting $t_0 = s/v_0$ and (3),

\langle v\rangle = \frac{2s}{\dfrac{s}{v_0}+\dfrac{2s}{v_1+v_2}} = \frac{2}{\dfrac{1}{v_0}+\dfrac{2}{v_1+v_2}}. \tag{5}

Clearing the two denominators,

\langle v\rangle = \dfrac{2v_0(v_1+v_2)}{2v_0+v_1+v_2}

(6)

Check

If $v_1=v_2=v_0$ the whole trip is covered at one constant speed, and the formula collapses to $\langle v\rangle = 2v_0\cdot 2v_0/(4v_0)=v_0$ , as it must. Also, $\langle v\rangle$ sits between $v_0$ and the average of $v_1,v_2$ , as a weighted mean of two legs should.

Remark

Taking $v_0\to\infty$ in the boxed result gives $\langle v\rangle\to v_1+v_2$ , not merely their average: an instantaneous first half consumes none of the time budget, which is then spent entirely on the second half, so the whole trip’s average velocity can exceed both $v_1$ and $v_2$ , a reminder that a time-weighted average of speeds is not the same object as a distance-weighted one.