Why isn't average speed equal to displacement over time?

I'm in an introductory Physics course and I need help!

During a one-hour trip, a small boat travels 80.0km north and then travels 60.0km east. What is the boat's average speed during the one-hour trip?

I found that the displacement is 100km. But I did not arrive at the right answer by dividing 100km/1hour. Why doesn't this give me the average speed?

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Hi Brijette, and welcome to Physics Stack Exchange! I edited your question a bit to make it clear that you're asking about a concept, not looking for an answer (since the latter is not allowed here). –  David Z Feb 18 '13 at 20:46
What if it had come back to the start point? What would be the average speed by your definition? –  Martin Beckett Feb 18 '13 at 20:51
I know that the answer is 140 km/h. I just don't see how? –  graybrij Feb 18 '13 at 21:06
I see what you are saying Martin...I can't use displacement in determining speed. But then what do I use? –  graybrij Feb 18 '13 at 21:08
Damn, that's a very very fast boat! –  Plouf Feb 19 '13 at 9:57

You use the total amount of movement over time. So here that is|:

80km plus 60km equals 140km


Which gives you the correct answer.

Displacement, using Pythagoras, would be 100km, but you travelled 140km in that hour! You didn't travel along that hypoteneuse, so it is irrelevant here.

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+1 for answering the question in a way that's relatively easy to understand, considering that the OP is an introductory student. –  Kevin Feb 19 '13 at 7:31

The formula you wanted to use gives you the magnitude of the average velocity, not the average speed.

To get the magnitude of the average velocity, you take the total displacement (which is a vector!), divide by the total time, and find the magnitude of that vector. What you get is:

$$\text{Magnitude of Average Velocity}= \biggl| \frac{\sum_i \vec{d}_i} {\Delta t_{\text{total}}} \bigg|=\bigg| \frac{\sum_i \vec{v}_i \Delta t_i} {\Delta t_{\text{total}}} \bigg|$$ Where the $\vec{v}_i$ are the different velocities, $\Delta t_i$ are the amounts of time spent at each velocity, $\vec{d}_i$ are the individual displacements ,and $\Delta t_{\text{total}}$ is the total amount of time.

To get the average speed, you take the magnitude of the individual displacement vectors, then sum and average them, giving this formula: $$\text{Average Speed=} \frac{\sum_i |\vec{d}_i|} {\Delta t_{\text{total}}} = \frac{\sum_i |\vec{v}_i| \Delta t_i} {\Delta t_{\text{total}}}$$

The difference is the order in which you take the absolute value and do the sum. This makes a big difference in some cases: the average velocity of helium atoms in a stationary balloon is $0$, but the average speed may be hundreds of $m/s$ (depending on the temperature).

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So you are looking for the average speed of a board during a one hour period. The total distance traveled by the boat is 140 km (80 + 60) in the course of an hour, giving the boat an average speed of 140 km/hr.

The confusion I feel in these types of problems, is from the 1 hour. When you get the answer merely by summing it dosent impart an understanding. The same problem with 2 hours instead of 1 may be clearer. In this case you have to sum the distances and divide by the time (which was done in the last problem but because you divided by 1 it had no effect.) So the answer in this case would be (80km + 60km)/2hr = 70 km/hr

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