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V-T Graphs - ROUND 2!

  1. Dec 19, 2006 #1
    1. The problem statement, all variables and given/known data

    distance (cm) 0.0 5.0 15.0 30.0 50.0 75.0 105.0 140.0
    time (s) 0.0 0.1 0.2 0.3 0.4 0.50 0.6 0.7

    a. calculating the displacement during each time interval. Add to the above table of data.

    b. calculating the velocity of the cart during each time interval. Add to the above table of data.

    c. sketching a velocity-time graph.

    d. calculating the acceleration of the object from the velocity-time graph.

    e. Using the velocity-time graph to calculate the total displacement of the cart.

    2. Relevant equations

    3. The attempt at a solution

    http://www.freewebs.com/dazed42/phys.htm [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Dec 19, 2006 #2


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    Staff Emeritus
    Science Advisor

    Ok, you've used the right method, and therefore the answer will be correct as long as you've not made any arithmetical errors. The reason for the displacement not equalling the total distance measured will be the same as the previous question, namely that the velocity of the truck at t=0 was no zero.

    So, you can use the equation given in the last question to calculate the initial velocity. Recall s=ut+(at^2)/2. Put s=140, t is the total time (0.7s) and a is the acceleration you have calculated.

    Really, for the question, you just need to state the the truck was not at rest at t=0, but if you want to check, you can calculate u from the above equation, then add it to each value of v on the graph, giving you a new graph, for which the area underneath will add up to 140cm
  4. Dec 19, 2006 #3
    Thanks buddy. From the phrasing of the question I was given the impression that the initial velocity could not be zero. (the cart is at rest before 0 and released at 0, so there must be some instantaneous acceleration)

    Much thanks for all the help, I'll be glad to put this behind me now. (Until finals.. whee)

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