Avg Velocity and Avg Speed

  • #1
badtwistoffate
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doing this problem 400km East in .75 hrs from a to b and 960 km south in 1.5 hrs from b to c.
Ang Velocity is /Delta x divided by /delta t correct?
and avg speed is just adding the distance's and times and diving by the time right?
Well for the avg velocity for this problem its saying its wrong any help?
 

Answers and Replies

  • #2
Doc Al
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Show exactly what you did. To find the average velocity, start by finding the displacement.
 
  • #3
badtwistoffate
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found displacement by using a right-hand coordinate system. and did solved by doing a^2 + b^2 = c^2. got the displacement to be 1040 km for the total trip. (Checked it its right)
Did velocity avg by 960-400/1.5-.75 = 747 and it says its wrong...
 
  • #4
Doc Al
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badtwistoffate said:
found displacement by using a right-hand coordinate system. and did solved by doing a^2 + b^2 = c^2. got the displacement to be 1040 km for the total trip. (Checked it its right)
You found the correct displacement, now use it to find the average velocity.
Did velocity avg by 960-400/1.5-.75 = 747 and it says its wrong...
I have no idea what you are calculating here. What happened to the displacement that you just calculated? That's what you need to use.
 
  • #5
badtwistoffate
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well it said change in x over change in t
960 and 400 was the two distances, how to apply the displacement? do i subtract from that?!
 
  • #6
Doc Al
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Average velocity is just displacement divided by time. Do it!
 
  • #7
badtwistoffate
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but what does my schaums outline says change in x over change in timee, I am confused :(
is it displacement over change in time or total time?

also isn't displacement delta x?
 
Last edited:
  • #8
Doc Al
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Displacement is measured from origin to final position; so is the time. Average velocity = the total displacement divided by the total time. The "change" in time is from 0 to the final time.
 

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