I have a super simple question on average velocity that either I am not setting up correctly, or is itself graded incorrectly in a huge online system used by thousands of students for a very long time.

I seriously doubt the later could go uncaught.

Here's the question:

A car travels along a straight line at a constant speed of 40.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. The average velocity for the entire trip is 31.5 mi/h.

What is the constant speed with which the car moved during the second distance d?

And here's my work:

V0 = 40.0 mph
Δx0 = d

V1 = ?
Δx1 = d

Vavg = 31.5 mph = ( 40.0 mph + V1 ) / 2
63 mph = 40.0 mph + V1
63 mph - 40.0 mph = V1
23 mph = V1

I've done this calculation a few times, used multiple sources to verify Vavg = ( Vf - Vi ) / 2 for constant acceleration, and even tried a few variations of the problem using different values from my book.

Stil no dice; the computer always marks my answers as "off by less than 10%".

What am I doing wrong?


closed as off-topic by tpg2114, user10851, user36790, Gert, John Rennie Jan 16 '16 at 7:26

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  • $\begingroup$ @AccidentalFourierTransform - Good point. $\endgroup$ – StudentsTea Jan 15 '16 at 19:58

The average velocity is given by $$ \bar v=\frac{1}{T}\int_0^T v(t)\mathrm dt=\frac{1}{T}(v_1t_1+v_2t_2) $$ where $t_1$ is the time spend on the first interval, $t_2$ is the time spend on the second one, and $T=t_1+t_2$.

Using $$ v_1t_1=v_2t_2=d $$ you get $$ \bar v=2\frac{v_1v_2}{v_1+v_2} $$

I believe you can take it from here.

  • 3
    $\begingroup$ Note that you cannot use $\bar v=\frac{v_1+v_2}{2}$. For example, if in your problem the car drives for 1 hour at $20\ \mathrm{mi/h}$, and then just one second at $40\ \mathrm{mi/h}$, then $\frac{v_1+v_2}{2}=30\ \mathrm{mi/h}$, which is not the average velocity (which is actually vero close to $20\ \mathrm{mi/h}$). The formula $\frac{v_1+v_2}{2}$ is average over distance, but what you want to calculate is average over time. $\endgroup$ – AccidentalFourierTransform Jan 15 '16 at 20:07

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