Average speed problem What am I missing?

[xXOfNiRXx]

Homework Statement

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?

Homework Equations

Average speed is equal to d/ change in t.

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.

 

[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.

 

[xXOfNiRXx]

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

 

[rl.bhat]

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

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

 

[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.

 

[rl.bhat]

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.

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

 

[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!