Two cyclists reaching the same finish

Problem

Two cyclists start from the same point, travel the same road and arrive at the same instant at the same finish. Can we conclude that they had the same instantaneous velocity at every moment? And the same average velocity? Justify.

Solution

Starting distinction. Two concepts must be kept separate:

• the average velocity depends only on the total distance and total time, i.e. on the endpoints of the trip;
• the instantaneous velocity describes the motion at each single instant and can vary freely along the way.

Same average velocity: yes. Both cyclists start from the same point and arrive at the same finish (same displacement $\Delta s$) taking the same time (same $\Delta t$, because they start and arrive together). Therefore

$\bar v = \frac{\Delta s}{\Delta t}$

is identical for both. The average velocity is necessarily the same.

Same instantaneous velocity: no. Having the same $\Delta s/\Delta t$ imposes nothing on how the distance is covered instant by instant. One cyclist may start slowly and accelerate at the end, the other sprint immediately and then slow down: completely different velocity profiles can still produce the same displacement in the same time. Indeed, if the profiles differ, at some points one will be faster and at others slower, and in general they will overtake each other more than once.

Conclusion.

$\ev{\text{Same average velocity: yes} \; \Rightarrow \; \frac{\Delta s}{\Delta t}\ \text{equal}; \qquad \text{same instantaneous velocity: not necessarily}}$

A useful remark still holds: since the two average velocities coincide, there is at least one instant at which the instantaneous velocities are equal (the $v(t)$ curves must cross), but this does not mean they are equal “at every moment”.

Collegamenti

Argomenti: Kinematics Concetti: Average velocity