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Homework Help: Graphical methods - dynamics problem.

  1. Apr 10, 2007 #1
    1. The problem statement, all variables and given/known data

    A car and a van are at rest on a straight horizontal road, with the van 25m in front of the car. At time t=0 seconds, the van moves off with an acceleration of 1.5 m/s². At time t=5 seconds, the car moves off in the same direction along the road with an acceleration of 2 m/s². use the sketches (I did a computer sketch and it can be found http://i16.photobucket.com/albums/b21/the_panic_light/carvan.jpg" [Broken]) of the velocity/time graphs of the car and van, on the same set of axes, to calculate the time at which their velocities are the same and state this velocity. Also, from the sketch, determine the distance between the van and car at this time.

    2. Relevant equations

    Well, because it's graphical methods, I've used v - b = m (t - a) with (a,b) = (5,0)

    The van's equation is v = 1.5t
    The car's equation is v = 2t - 10

    3. The attempt at a solution

    I equated the two equations to get the velocity being the same at t = 20, and substituted this in the van's equation to get that velocity to be v= 30. My main problem is with finding the distance between the van and the car at that time. The answer tell me it's 100m, but I've no idea what the method is. Please help! :cry: Thanks.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 10, 2007 #2

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    Do you know the relation between the distance and time given a constant acceleration?
  4. Apr 10, 2007 #3
    Well, um...no. :S

    The point of the exercise was to realise that the area under a certain line on a velocity/time graph equals the distance travelled. I then assumed that if I found the distance the van had travelled, and subtracted from it the distance the car had travelled, I would have my answer. But the answer I got that way did not equal the answer in the book...
  5. Apr 10, 2007 #4
    looking at the graph, it seems like the headstart is 15 seconds, not 5 as stated in the problem.
  6. Apr 11, 2007 #5
    Sorry, it's not to scale.
  7. Apr 11, 2007 #6
    oh, ok. What answer did you get, I'm looking at 75, using your graph.

    oops, that should be 100, the van had a horizontal offset. (one of those pesky integration constants)
    Last edited: Apr 11, 2007
  8. Apr 11, 2007 #7
    The answer's meant to be 100...but again, I'm not meant to be using integration. I'm meant to be using the area under each graph as the distance travelled...so looking at the graph, before reaching t=20, the van travelled:

    the area under the blue line between t=0 and t=20: 1/2 x 20 x 30 = 300m

    and the car travelled:

    the area under the green line between t=0 and t=20: 1/2 x 15 x 30 = 225m

    ...but how does this help me get to that 100m apart?!
  9. Apr 11, 2007 #8
    no i understand that your're not supposed to be using integration at least in the formal sense, tho thats exactly what you are doing by computing area.

    your solution above is correct, except that it ignores the initial offest between the van and the car. (this is what I meant by constant of integration--sometimes these get lost and mess up results)
  10. Apr 11, 2007 #9
    But by integrating between t=0 and t=20 for the van (as it starts from 0) and only integrating between t=5 and t=20 for the car, surely I'm taking the initial offset into consideration?
  11. Apr 11, 2007 #10
    reread the problem carefully. It states that the van starts from a position 25 ahead of the car. You have taken into account the offset by virtue of the headstart, but not the physical offset that has nothing to do with time.
  12. Apr 11, 2007 #11
    OH. Oh my God, that's so obvious. Thanks so much for pointing that out! So I am actually right, 75 from the areas + the initial 25m offset = 100m. Got it. Thank you!
  13. Apr 11, 2007 #12
    No sweat, it is there to teach you about those pesky constants that get lost I suspect. And yes you are right!
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