Runner's Average Velocity: 0.014 m/s

In summary, the runner's average velocity for the entire journey from start to finish is 0.476 m/s west. This is calculated by dividing the displacement of 300 meters by the total time of 630 seconds. The runner first moves 1200 meters east at a velocity of 4.0 m/s for 5 minutes (300 seconds), then moves 1500 meters west at a velocity of 5.0 m/s for 5 minutes. The distance traveled is greater going west because the runner's velocity is greater.
  • #1
9
0

Homework Statement


A runner moves with a velocity of 4.0 m/s due east for 5.0 minutes. The runner then rests for 30 seconds, turns around and heads due west with a velocity of 5.0 m/s for 5 minutes. What is the runner's average velocity for the entire journey from start to finish?

Homework Equations


Is there a different formula for velocity and average velocity?

The Attempt at a Solution


distance/time 9/630=0.014
 
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  • #2
Mary27 said:
distance/time
No, distance/time gives average speed. Velocity is a vector, and average velocity is displacement/time.
What is the vector displacement for the journey?
 
  • #3
Is the vector displacement -1
Does the rest period for 30 seconds come into the equation?
 
  • #4
Mary27 said:
Is the vector displacement -1
No, to the above question.

Yes, to the following question.
Does the rest period for 30 seconds come into the equation?
Displacement is the straight line distance from the starting point to the ending point.
 
  • #5
SammyS said:
Displacement is the straight line distance from the starting point to the ending point.
Small, ahem, clarification. Displacement is a vector; its magnitude is the straight line distance from the starting point to the ending point.
 
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  • #6
Mary27 said:
Is the vector displacement -1
 
  • #7
So then is the total displacement 4+5?
And the time would be 630 seconds?

The equation would be: 9/630= 0.0142857
 
  • #8
Mary27 said:
So then is the total displacement 4+5?
There are at least two problems with that calculation.

If you go at 4 meters per second eastward, how far will you have moved from the starting point after 5 minutes?

If you then go at 5 meters per second westward for another 5 minutes, how far will you be from the original starting point? And in which direction?
 
  • #9
4m/s x 300sec=1200m/s
5m/s x 300sec=1500m/s
1500m/s- 1200m/s=300m/s
300m/s /630=.476m/s
Is the direction west or east? I think west.
 
  • #10
Mary27 said:
4m/s x 300sec=1200m/s
Units? If you multiply meters per second times seconds you get what?
Mary27 said:
Is the direction west or east? I think east.
Explain your reasoning.
 
  • #11
The first calculation is 1200m/s east
The second calculation is 1500m/s west
The distance is greater going west.
 
  • #12
Mary27 said:
The first calculation is 1200m/s east
Incorrect. Units.
The second calculation is 1500m/s west
Incorrect. Units.
The distance is greater going west.
What distance? You have not computed any distances.
 
  • #13
1500m/s-1200m/s=300m/s

300m/s / 630s=0.476m/s west
 
  • #14
Units!

Mary27 said:
1500m/s-1200m/s=300m/s
Nothing is moving at 1500 meters per second.
300m/s / 630s=0.476m/s west
If you divide 300 m/s by 630 seconds you do not get a velocity. You get an acceleration.
 
  • #15
I changed 5 minutes to 300 seconds and multiplied that by 5 m/s to get 1500.
 
  • #16
Mary27 said:
I changed 5 minutes to 300 seconds and multiplied that by 5 m/s to get 1500.
Units, units, units.

If you multiply 5 meters per second by 300 seconds, what units should the result have?
 
  • #17
meters per second
 
  • #18
Mary27 said:
meters per second

You have a severe problem with carrying the correct units or a severe problem with algebra, and this could be fatal for the rest of your class.

x multiply by y gives you "xy". The same goes with units. If you multiply something with units m/s with something with units "s", you get something with units "m". It is basic algebra, i.e.

m/s * s = m

So that product gives you something that is a length!

Zz.
 
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