Average Velocity

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  • #1
sfgradv
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Does anyone know how to do this? Thanks!
 

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  • #2
quasar987
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Yes I do, and you're welcome.

On this forum, you are expected to show your work so we can help you progress. We will not give you the answer. SO, what have you done so far? Can you at least tell me what is the equation for the average velocity?
 
  • #3
sfgradv
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delta v over delta t. i don't no the velocity for t = 5.
 
  • #4
electro05
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hello I am guessing since ave velocity is m/s you can get the velocity by looking at the y-axis aka the position in meters. Then you have the components for solving the problem, right? Others-feel free to correct me if I'm wrong.
 
  • #5
sfgradv
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the thing is it doesn't tell u exactly what the position is at t = 5. it is between 40 - 60 and it said not to eyeball it. i thought it was 45 and i used it in the formula and got 6.3 and got the wrong answer. so what should i do now?
 
  • #6
jtbell
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sfgradv said:
delta v over delta t.

That's the formula for average acceleration. Average velocity is [itex]\Delta x / \Delta t[/itex].
 
  • #7
sfgradv
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yea sorry. i meant delta x over delta. i still don't no how to do it and its 3 am now! i have to go to sleep. ahhhhh. can someone please help me quick?
 
  • #8
jtbell
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sfgradv said:
the thing is it doesn't tell u exactly what the position is at t = 5. it is between 40 - 60 and it said not to eyeball it. i thought it was 45 and i used it in the formula and got 6.3 and got the wrong answer. so what should i do now?

Get out a ruler, and carefully measure (a) the distance between the x = 40 and x = 60 lines on the graph, and (b) the distance between the x = 40 line and the point where the graph intersects the t = 5 line. That will give you enough information to calculate x at t = 5 (more precisely, x - 40).
 
  • #9
HallsofIvy
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Don't worry about t=5. Since the graph is a straight line, the velocity is a constant there. It looks to me like t= 4.5, x= 60 and when t= 6, x= 20. velocity at t= 5 should be (20- 60)/(6-4.5) (which is no where NEAR 6.3!)
 

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