average velocity

[kimikims]

I know average velocity is V = displacement/time, but how do you find this with 3 different sets of numbers??
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A car travels along a straight stretch of road. It proceeds for 13.9 mi at 56 mi/h, then 29 mi at 42mi/h, and finally 32.2 mi at 35.2 mi/h. What is the car's average velocity during the entire trip? Answer in units of mi/h.

[PRodQuanta]

V_{ave}= \frac{d_{total}}{t_{total}

Paden Roder

[PRodQuanta]

Well, for some reason that didn't work.

Ave Velocity= total distance/total time.

Paden Roder

[kimikims]

I got (13.9 + 29 + 32.2) / (56 + 42 + 35.2) = .5638138138

That's wrong, what did I do wrong??


Well, for some reason that didn't work.

Ave Velocity= total distance/total time.

Paden Roder

[recon]

I got (13.9 + 29 + 32.2) / (56 + 42 + 35.2) = .5638138138

That's wrong, what did I do wrong??

You are dividing the total distance by the sum of all the different speeds the car travelled at. This is wrong. You need to divide total distance travelled by the total time. To find the total time, you need to find the time travelled for each of the different speeds, then add them all up. Remember that time = Distance/Speed.

Actually, I find this question kind of weird. To have travelled at different (decreasing) speeds, the car must have been decelerating. PRodQuanta's solution does not apply to real life (neither does the question) because the question assumes that the car made the transition from one speed to another instantaneously.

[arunma]

Actually, I find this question kind of weird. To have travelled at different (decreasing) speeds, the car must have been decelerating. PRodQuanta's solution does not apply to real life (neither does the question) because the question assumes that the car made the transition from one speed to another instantaneously.
Yeah, I guess only us physics people would notice that. I'm guessing kimi is taking an algebra based physics course. And if they talk about decelerations, then the problem gets a bit messier. I guess that since the distances are large, the author's intended calculation is still a very good approximation.