Cyclist over three legs

Problem

A cyclist covers three consecutive legs, as in the figure: $100\;\text{m}$ at constant velocity in $5\;\text{s}$, then $150\;\text{m}$ accelerating uniformly in $10\;\text{s}$, then $250\;\text{m}$ again at constant velocity in $15\;\text{s}$. Calculate the average velocity over the whole route and estimate the maximum velocity reached.

Solution

Average velocity over the whole route. The average velocity is, by definition, the total distance divided by the total time. It is not the average of the velocities of the individual legs.

$s_{\text{tot}} = 100 + 150 + 250 = 500\;\text{m}, \qquad t_{\text{tot}} = 5 + 10 + 15 = 30\;\text{s}$

$\bar v = \frac{s_{\text{tot}}}{t_{\text{tot}}} = \frac{500}{30} = 16{.}7\;\text{m/s}$

$\ev{\bar v = 16{.}7\;\mathrm{m/s} \approx 60\;\mathrm{km/h}}$

Estimate of the maximum velocity. The instantaneous velocity is highest in the middle leg, the one in which the cyclist accelerates: it starts from the velocity of the first leg and grows up to the peak value, reached at the end of the acceleration. Let’s compute the reference velocities of the constant-velocity legs:

$v_1 = \frac{100}{5} = 20\;\text{m/s}, \qquad v_3 = \frac{250}{15} = 16{.}7\;\text{m/s}$

In the accelerated leg the velocity climbs above these cruising values before settling back to the third leg’s regime: the peak sits a little above them, with a reasonable estimate of

$\ev{v_{\max} \approx 22\;\mathrm{m/s} \approx 79\;\mathrm{km/h}}$

This is an estimate: the precise figure would require knowing the instant-by-instant velocity profile in the middle leg, whereas here we only have the total distance and time of each leg.

Collegamenti

Argomenti: Kinematics Concetti: Average velocity · Uniformly accelerated motion Competenze: Graph reading