Calculate your average velocity from pillar to post

[Heat]

Homework Statement

Starting from a pillar, you run a distance 200m east (the x-direction) at an average speed of 5.0m/s , and then run a distance 280m west at an average speed of 4.0m/s to a post.

Calculate your average speed from pillar to post.

Calculate your average velocity from pillar to post.

I drew a rough sketch on how I would this would look like:
http://img365.imageshack.us/img365/449/untitled1zs5.jpg

I understand that they are on the sime line, as if the runner is running forward and back,
I drew it in two separate lines for convienience.

Homework Equations

V = X2-X/T2-T1

The Attempt at a Solution

At first, I would assume that on they way back, the speed would be 4.0m/s and would remain that for the 80m portion gap between 200 and 280m.

Then I decided, that was too easy (and plus I got it wrong), so I think the equation shown above comes into play. But I don't know where to go from there, can anyone stir me into the right direction.

 

[learningphysics]

Average speed and average velocity are different.

Average speed = (Total Distance)/(Total Time)

Find the total distance... find the total time for the trip... divide them for the average speed...

Average velocity = (Total Displacement)/(Total Time) (displacement is a vector)

 

[Heat]

Average speed and average velocity are different.

Average speed = (Total Distance)/(Total Time)

Find the total distance... find the total time for the trip... divide them for the average speed...

Average velocity = (Total Displacement)/(Total Time) (displacement is a vector)

Thank you for your fast response.

I understand that Average speed is total distance over total time, which leads the av. speed to be 4.4 m/s.

For Average velocity, you mention that it's total displacement over total time.

The displacement would have to be 80m, -80m since it's going 80m from where it started.
Total time is where I am starting to have problems,

I know that the 200m @ 5m/s has a total time of 40 seconds.
I know that the 280m @ 4m/s has a total time of 70 seconds.

Would total time be 30 seconds? or will be still be 70+40= 110s.

Average Velocity = -80m/30s
or
Average velocity = -80m/110s?

 

[learningphysics]

Average speed looks right. Total time is 110s. So average velocity would be -80/110 m/s. And since east is positive and west is negative, in your final answer I would use 0.727 m/s West.

 

[Heat]

Average speed looks right. Total time is 110s. So average velocity would be -80/110 m/s. And since east is positive and west is negative, in your final answer I would use 0.727 m/s West.

Thank you, I was thinking about .727, but thought to myself that the number might be too small for average velocity.

I understand that -80m is the displacement, and time is 110 as you mentioned total time. Time cannot be negative, so the final result would have to be -.727 m/s.

I thank you again for the time you took to help me out.

 

[learningphysics]

Thank you, I was thinking about .727, but thought to myself that the number might be too small for average velocity.

I understand that -80m is the displacement, and time is 110 as you mentioned total time. Time cannot be negative, so the final result would have to be -.727 m/s.

I thank you again for the time you took to help me out.

No prob. be sure at the end to interpret the negative sign as being west... ie: writing the final answer as 0.727m/s West.