# Not exactly a homework issue - just trying to find a formula

mrgenius24
Alright, first I should probably let you all know that this is very simple physics, so I'm just looking for a quick answer.

1. Homework Statement

I'm trying to find a simple way to calculate the average velocity of an object where we know the average velocity during a number of segments of the distance traveled and we know that the segments of distance are all equal but we do not know the time elapsed. I would also like to know the opposite - what if we know of a couple of equal segments of time but we know nothing about the total distance traveled. I think that I know how to solve the latter, but I am not sure. I'll give you two examples so you know what I mean, in case that I was not clear enough:
1. A car traveled a certain distance during a certain amount of time. For 1/3 of the total distance, it traveled at 10 m/s. For the next 1/3, it traveled at 9 m/s, and for the final 1/3, it traveled at 15 m/s. What's the average velocity?
2. A car traveled a certain distance during a certain amount of time. For 1/3 of the total time, it traveled at 10 m/s. For the next 1/3, it traveled at 9 m/s, and for the final 1/3, it traveled at 15 m/s. What's the average velocity?

Also what happens if we have a problem like this:
3. A car traveled a certain distance during a certain amount of time. For 1/2 of the total distance, it traveled at 10 m/s. For 1/2 of the remaining time, it traveled at 9 m/s, and for the next 1/2 of the remaining time it traveled at 15 m/s.
What formula do we use then?

## Homework Equations

Of course, I know of the formula which states that average velocity equals total distance traveled divided by the total time elapsed. I also think that I know that when we know that when the segments of time are equal, like in the example problem I gave above, we can just add together all the velocities and divide them by the total amount of time segments like this: (10 m/s + 9 m/s + 15 m/s)/3.

## The Attempt at a Solution

I know how to solve this problem the long way - I am just looking for the quick formula I was once told by a teacher but I can't remember it now.

If you know the velocity and the distance over each segment, you can find the time it took to traverse each segment. Sum up the times over each of these segments and you will get total time for the trip. You now know the total time and the total distance, meaning you can find the average velocity for the entire trip.

I also think that I know that when we know that when the segments of time are equal, like in the example problem I gave above, we can just add together all the velocities and divide them by the total amount of time segments like this: (10 m/s + 9 m/s + 15 m/s)/3.
Why do you think this works? Try it the correct way. Does this give you the same answer?

mrgenius24
If you know the velocity and the distance over each segment, you can find the time it took to traverse each segment. Sum up the times over each of these segments and you will get total time for the trip. You now know the total time and the total distance, meaning you can find the average velocity for the entire trip.
Hi mate, first of all I appreciate the answer.

Perhaps I was not clear enough, I meant that the exact distance is unknown. All we know is this:
The car has traveled the first third of the trip at 10 m/s, the second third at 9 m/s, and the final third at 15 m/s. It is unknown how long the distances are exactly - it is only known that they are all equal. They could all be 100 m or 100 km.

Why do you think this works? Try it the correct way. Does this give you the same answer?
I have just tried it, and it seems that it does in fact work.

So, let's say a car travels at 10 m/s for 30 seconds. It's going to pass 300 m - so s1 = 300m, t1 = 10s. It travels at 20 m/s for another 30 seconds - it's going to pass 600 m. So, s2 = 600m, t2 = t1 = 10s. And finally, it travels at 30 m/s for another 30 seconds - so it's going to pas 900 m - so s3 = 900m, t3 = t2 = t1 = 10 s.

So let me try to do it properly: (s1+s2+s3)/(t1+t2+t3) = 1800m/90s = 20m/s.
And then I'll do it like this: (v1+v2+v3)/3 = 60m/s/3 = 20m/s.
It works with the 10m/s, 9m/s and 15m/s scenario too.

It works with the 10m/s, 9m/s and 15m/s scenario too.
Do you mind showing what you did for this one specifically then? Because, if you've done it correctly, it doesn't give you the same answer like you're saying it does. :P

haruspex
Homework Helper
Gold Member
2020 Award
It is unknown how long the distances are exactly -
So let the distances all be x. What happens when you RedDelicious' approach?

mjc123
Homework Helper
Do you mind showing what you did for this one specifically then? Because, if you've done it correctly, it doesn't give you the same answer like you're saying it does. :P
In this instance it does. When the time intervals are equal, the average speed is the arithmetic mean of the segment speeds. When the distance intervals are the same, the average speed is the harmonic mean of the segment speeds, i.e. 1/V = (1/n)*Σ(1/Vi). When the intervals are not all equal in time or distance, you have to do it the long way.

In this instance it does. When the time intervals are equal, the average speed is the arithmetic mean of the segment speeds. When the distance intervals are the same, the average speed is the harmonic mean of the segment speeds, i.e. 1/V = (1/n)*Σ(1/Vi). When the intervals are not all equal in time or distance, you have to do it the long way.
I have to disagree with you. He claimed that, for the original problem, the average velocity for the trip is (15+9+10)/3 and that you get the same answer when you do it the "long way." I asked him to try it both ways and see if that was actually true. He responded again with yes it works and that you get the same answer. That post with equal time intervals was his justification. I still stand by that being wrong. In the original problem, the time intervals are not equal and so the average speed is not just the arithmetic mean. When you break it up into segments to find the total trip time, and then find the average speed that way, as I suggested he do in my first post, you get the correct answer.

haruspex