# Average Velocity

1. Aug 31, 2003

### ffrpg

Here's the problem, While Traveling through the first half of the distance of the trip, your average velocity is 14.0 m/s. What average velocity would you need during the second half of the distance on your trip, if you wanted your average velocity for the entire trip to be 24.0 m/s.

Here's what I did. I used the formula V_av = 1/2(v_o+v). I plugged in 24.0 m/s into V_av and plugged in 14.0 m/s into V_o. I solved from there and got 34.0 m/s. That isn't the right answer. Did I use the right formula? Is there a way to express the time for the first and second half of the trip?

2. Aug 31, 2003

### HallsofIvy

No, you didn't use the right formula. The idea of any "average" (and there are many different kinds) is that you can use the average in place of al the original numbers. The "arithmetic average" is a single number by which you can replace all the other numbers and get the same SUM- that's the one that gives (x+y)/2 as the average of x and y. (If you were concerned with products, you would use the geometric average.)

Since you are talking about speed, the "average" speed for the entire trip is the speed such that, if you traveled at that speed only, you would cover the same distance as you actually did.

In this problem you are told that the average speed for the first half of the trip is entire trip is 14 m/h. You want the average for the entire trip to be 24 m/h. What must your average speed be for the entire trip.

Assume, for now, that the entire trip covers 240 mi. The first half, then, is 120 miles. At 14 m/h, that takes 120/(14 m/h)= 60/7= 8 and 4/7 hours. You want to average 24 m/h for the entire trip. Okay, that means the entire trip must take 240/24= 10 hours. You have 10- 8 4/7= 1 3/7= 10/7 hours in which to cover the second 120 miles: that's an average of 120/(10/7)= 12*7= 84 m/h!!!

Notice I said "assume 240 mi." How do we know the answer wouldn't be different if the entire trip were a different distance? We don't until we check. The point of using 240 mi (which I chose simply because it was easily divisible by 24) was that it helped to see what operations we needed to do.

Now suppose the entire trip is A mi. Then half of it is A/2 and, at 14 mph, that will require (A/2)/(14)= A/28 hours. If we average 24 m/h for the entire distance, that will require A/24 hours.
That means we have A/24- A/28= (7A- 6A)/(4*6*7)= A/168 hours in which to do the second half. We have to average (A/2)/(A/168)= 84 m/h.

Yes, that does NOT depend upon A and is the answer we got before. That is the average velocity required.