Example: Average velocity


Understanding the situation

Although the concept of average velocity seems straightforward when described in words -- average velocity is the total displacement divided by the amount of time to make that displacement -- it can be quite tricky in practice. Part of the reason is that depending on whether distances or times come to the fore, our intuitions can be led astray. Let's try a sample problem as an example.

Presenting a sample problem

Source: Patrick Bell, 2007; Wikimedia CommonsIn a 4x100 m relay race, each of four runners take turns running a 100 m course, trading places by passing a baton as shown in the figure at the right. In one specific example of a relay race between two teams, the speeds of the runners in the one of the teams are graphed below. (To simplify the problem in order to make it do-able during an exam, we are ignoring the short amounts of time it takes the runners to get up to speed and the small variations in their speed during their 100 m lap.) Find the average speeds of the individual runners and the team’s average speed.

Solving this problem

The first part of the problem -- find the average speeds of the individual runners -- is easy. Each runner runs at a constant speed (ignoring the speed-up and slow-down times) so there speeds are equal to their average speeds, which we can read off the graph.

    • Arnie     8.0 m/s
    • Ben        6.0 m/s
    • Jack       7.0 m/s
    • Phil        8.5 m/s 

The average speed of the team is a little trickier. You cannot just average the averages. Since they each ran for different times, some speeds contribute more than others. What matters is the total distance divided by the total time since $\langle v\rangle = Δx/Δt$.

Now we could take a shortcut and assume that each runner ran a total distance of 100 m. Then the total distance would be 400 m and the total time, as read off the graph, would be 56 seconds. This would give an average speed

<$v$> = (400 m)/(56 s) = 7.1 m/s.

Note that the averages of the velocities: (8.0 m/s + 6.0 m/s + 7.0 m/s + 8.5 m/s)/4 = (29.5 m/s)/4 = 7.4 m/s, is NOT the same as the average velocity. This might not seem like much of a difference, but in 56 seconds, an extra 0.3 m/s would correspond to a distance of nearly 17 meters - a huge gap in a race like this.

But this assumes that this team won and therefore made it to the finish line. If another team had won, this team might have stopped short. In fact, the graph might not be correctly drawn! We really ought to check. Did each runner indeed run a distance of 100 meters? 

It's harder to read the times off the graph than the velocity, so lets use the velocities and check the times. For each runner, if they had run 100 meters, we would expect their times to be given by $Δt = Δx/\langle v\rangle$. For each runner this would be

  • Arnie     Δ$t$ = (100 m)/(8.0 m/s) = 12.5 s
  • Ben        Δ$t$ = (100 m)/(6.0 m/s) = 16.7 s
  • Jack       Δ$t$ = (100 m)/(7.0 m/s) = 14.3 s
  • Phil        Δ$t$ = (100 m)/(8.5 m/s) = 11.8 s

These times look right on the graph so we can infer that our result is correct. If, for example, Phil had only run for 10.0 s, then his distance would have been  $Δx = \langle v \rangle Δt$ = (8.5 m/s) x (10.0 s) = 85 m and his team would have lost by 15 meters.

Joe Redish 12/30/14

Article 327
Last Modified: February 24, 2019