average speed

[pizza1994]

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A car drives to a location with an average speed of 30 m/s and makes the return trip with an average speed of 40 m/s. What is the magnitude of the car’s average speed for the entire trip?

the answer is 34 m/s according to TPR!

how???? can someone explain what is happening? 🙁 Thanks so much!

 

[el_duderino]

My answer would be 0 m/s! I guess that would be the average velocity, though.

Anyway, let's say the location is x meters away. d = rt, so x = 30(tout), tout = x/30 and tback = x/40

ttotal = tout + tback and dtotal = r(ttotal) = 2x

2x = r(tout+tback) = r(x/30+x/40)
2x = rx/30 + rx/40
2 = r/30 + r/40
240 = 4r + 3r = 7r
240/7 = r = 34.28 m/s

Or the quick way:

Assume the place is 120 meters away. Trip out takes 4 seconds, trip back takes 3 seconds. That's 240 meters in 7 seconds or 240/7 = 34.28 m/s

 

[Lunasly]

My answer would be 0 m/s! I guess that would be the average velocity, though.

Anyway, let's say the location is x meters away. d = rt, so x = 30(tout), tout = x/30 and tback = x/40

ttotal = tout + tback and dtotal = r(ttotal) = 2x

2x = r(tout+tback) = r(x/30+x/40)
2x = rx/30 + rx/40
2 = r/30 + r/40
240 = 4r + 3r = 7r
240/7 = r = 34.28 m/s

Or the quick way:

Assume the place is 120 meters away. Trip out takes 4 seconds, trip back takes 3 seconds. That's 240 meters in 7 seconds or 240/7 = 34.28 m/s

This seems overly complicated. Shouldn't we just take the average of the two speeds given? (30 + 40) = 35 m/s.

 

[el_duderino]

This seems overly complicated. Shouldn't we just take the average of the two speeds given? (30 + 40) = 35 m/s.

Yes, you can do that, if you want to be completely wrong. 35 != 34.28 is it?

How about the average of 50 and 100 m/s there and back? Must be 75 m/s average speed right? Nope. It's actually 67 m/s.

 

[Lunasly]

Lol, relax.

So, why is this the case? Why do we have to make up an arbitrary distance (e.g., 120 m)? I'd assume it is because we are talking about a definite distance that takes a given amount of time to travel depending on the speed at which the object is moving.

Edit: Found this online:

"Conceptually speaking, just like the average test score is "weighted" toward the score that occurs most often, average speed is weighted toward the portion of the journey that takes the longest. Since the 30 m/s segment takes the longest, it should count more than the 40 m/s segment. The average speed, therefore, should be less than 35 m/s. Choice B = 34 m/s!"

 

[el_duderino]

Lol, relax.

So, why is this the case? Why do we have to make up an arbitrary distance (e.g., 120 m)?

We don't, it just makes the calculation faster because you don't have to carry a variable around the whole equation. I did it both ways, if you noticed.

The question implies that the average speed is going to be the same regardless of the distance, so a trick is to assume any arbitrary distance and then do the calculations. We see when we do it more rigorously that the distance variable ends up canceling out.

 

[pizza1994]

We don't, it just makes the calculation faster because you don't have to carry a variable around the whole equation. I did it both ways, if you noticed.

The question implies that the average speed is going to be the same regardless of the distance, so a trick is to assume any arbitrary distance and then do the calculations. We see when we do it more rigorously that the distance variable ends up canceling out.

Perfect I get this now!!1 thanks so much!!! 🙂