Question about basic physics...

[alexjohnson16]

A runner runs 100 m on a straight track in 11s and then walks back in 80s. What are the average velocity and the average speed for each part of this motion and for the complete motion?


D=100 m D=100m
T= 11 s T=80s
V=? V=?
V=d/t V=d/t
V=100m/11s V=100m/80s
V=9.09 m/s V=1.25 m/s

D=200m
T=91s
V=?
V=d/t
V=200m/91s

I just did this and my friend tells me I'm wrong, can't explain how... But I think I'm right??? Any physics help here???

Thanks in advance...

 

[mugs]

80+11=91, not 93

 

[alexjohnson16]

Err yeah, sorry about that... But its hte general idea of how I did the problem that he says is wrong...

 

[mugs]

Is it possible that he's wrong? Because that looks right to me.

The only other thing I can think of is that velocity is speed with a direction... so his velocity coming back could be negative? In which case his average velocity would be 0, because his net change in position is 0m over 91s... but I doubt that's right.

 

[TechnoKid]

Velocity is an (speed) acceleration with direction.

edit: I should note that I would consider speed to be the magnatude of acceleration. I think i am confusing something here however. I'll check my book later. Its late.

 

[alexjohnson16]

Okay one more question, I missed the day on uniform acceleration and am unsure how to do these problems...

The problem is: A train traveling at 100km/h slows down with a uniform acceleration of -0.60 m/s². How long, in seconds, does it take to stop?

I just need to know the formulas used when dealing with these problems...

Thanks again...

 

[mugs]

Convert .60 m/s2 to km/h/s... multpy by 3600 (seconds per hour), divide by 1000 (m per km), or just multipy by 3.6. Then take that number and divide 100 by it. That should be the time in seconds that it takes to decelerate 100 km/h.

 

[alexjohnson16]

edit: thanks for hte help mugs :beer:

 

[mugs]

Speaking of Google...

And to clarify what TechnoKid was saying, speed is the magnitude (absolute value) of velocity. Re-reading your question, I see that the teacher asked for speed and velocity separately, which leads me to believe that he IS looking for a negative value in one direction.

 

[silverpig]

For the last part his average velocity is zero.

 

[TuxDave]

Originally posted by: silverpig
For the last part his average velocity is zero.

Bingo. If you want to find average velocity (not speed) you take the total displacement and divide by time. So for the complete loop, the person's final position is the same as the initial position, so his average velocity is zero.

And if you make his velocity positive in one direction, it must be negative in the opposite direction.

 

[TuxDave]

Originally posted by: TechnoKid
Velocity is an (speed) acceleration with direction.

edit: I should note that I would consider speed to be the magnatude of acceleration. I think i am confusing something here however. I'll check my book later. Its late.

No.

 

[Joker81]

Velocity = m/s
Acceleration = m/s^2



Don't Forget that you have to determine Velocity and Speed.