# Average speed problem What am I missing?

1. Aug 31, 2009

### xXOfNiRXx

1. The problem statement, all variables and given/known data
A person walks from point a to point b at a speed of 5 m/s. Then she turns around and walks from point b to point a at a speed of 3 m/s. What is her average speed?

2. Relevant equations
Average speed is equal to d/ change in t.

3. The attempt at a solution
So I know the distance is 2d. I also know the acceleration is constant, though I don't know if I need it for the problem. So I tried to do 5m/s + 3m/s /2 to get the answer. That led to the wrong path. Since we don't have time, I'm assuming that maybe I use a kinematic equation. However, we don't know what acceleration is, other than it is constant.

2. Aug 31, 2009

### rl.bhat

Here the person is walking with uniform velocity.
From a to b the time taken is t1 = d/5 m/s
From b to a the time taken is t2 = d/3 m/s
Find total time and total distance. Then the average velocity = total distance / total time.

3. Aug 31, 2009

### xXOfNiRXx

I see that total distance is 2, or 2d. I am having trouble seeing how to get the time.

4. Aug 31, 2009

### rl.bhat

Total time t = t1 + t1 = d/(5 m/s) + d/(3 m/s)

5. Aug 31, 2009

### xXOfNiRXx

Tried working it out, but I'm not seeing it here. I see that by manipulation you get t=d/v at which point you plug in the speed that I am given. Can you explain a bit more. I'm quite confused here.

6. Aug 31, 2009

### rl.bhat

Average velocity = 2d/t = 2d/( d/5 + d/3)

7. Aug 31, 2009

### xXOfNiRXx

Oooooooo!!! It all makes sense! Got it! Thanks so much! Really appreciate it. The trick was seeing that the d's would cancel! Thanks!

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