# Trouble with Simple Average Velocity [closed]

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 RennieJan 16 '16 at 7:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114, Community, Community, Gert, John Rennie
If this question can be reworded to fit the rules in the help center, please edit the question.

• @AccidentalFourierTransform - Good point. – 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 $$\bar v=2\frac{v_1v_2}{v_1+v_2}$$
• 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. – AccidentalFourierTransform Jan 15 '16 at 20:07