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Given position vs time graph, find velocity?

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Given the graph in the uploaded screenshot, what is the bicyclist's velocity at t=10s?


    2. Relevant equations

    I have been trying to do a slope equation? Rise/run.

    3. The attempt at a solution

    Tried to do slope equation. Looks like 75/10, which would be 7.5, but that answer is incorrect. Why? It wants to two sig figs, and that is two sig figs...


    I also have no clue what to do if t= anything else, like 25 or something. Is there a different equation for that? I'm so lost. :(
     

    Attached Files:

  2. jcsd
  3. Aug 23, 2011 #2
    rise over run wouldn't be 75/10. it starts at 50 and then moves to 75. so the change in distance is 75-50=25. so rise over run is 25/10
     
  4. Aug 23, 2011 #3
    remember speed is distance traveled/time taken. when time= 0, its position isn't zero like you assumed. It's position is 50. And then in 10 seconds its position is 75. So the distance traveled is 25. So s=d/t, where d=25 and t=10seconds.
     
  5. Aug 23, 2011 #4
    Would I do the same then, for an amount like 25 or 35? Or do I say 25 would be 0, since there is no slope there, and then 35 would be a negative number, because the slope is going downward?
     
  6. Aug 23, 2011 #5
    if t=25, all you need to do is look at the graph. what is happening at t=20? position is 100. now take t=25. position is still 100. what does that mean??? in 5 seconds the position of the cyclist has not changed which means he is stationary. not moving so velocity is zero. The formula can also be applied. s=d/t. between 20 and 25 seconds. distance traveled is zero in a time frame of 5 seconds. so s=0/5 = 0. which is the same as saying rise over run. Between 20 and 30 would be the same, distance travelled between 20 and 30 seconds is zero in a time frame of 10 seconds so s=0/10=0. At t=35, the slope has changed. at t= 30 seconds position was 100 but now at t=35, the position is 50 again. So distance traveled is final distance minus the initial distance so 50-100=-50. And that has occurred in a time frame of 5 seconds. so again s=d/t, -50/5=-10 so velocity is -10.

    That is why it has a negative gradient. So essentially, if the position at t=0 is 50 and then at t=25 is 100 and then at t=35 is 50 again it means he is going backwards. he has turned around at t=30 and is heading back
     
  7. Sep 19, 2011 #6
    thanks alot that help me alot too. I fully understand it now
     
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