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Homework Help: Quick Velocity Problem

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data

    When you are 20 m away from a stopped hummer it begins accelerating away from you at 3 m/s/s . With what constant velocity should you run to catch the hummer?

    2. Relevant equations

    Vf ^2 = Vi ^2 + 2ad

    d = Vi * t + 0.5at^2

    Vf = Vi + at

    3. The attempt at a solution

    I really don't know how to approach the problem. Assuming that they are asking for the minimum velocity...

    Distance for runner:

    Velocity (time) - 20 meters

    Distance for hummer:

    = 3/2 (t^2)

    To find the initial velocity you would set the two equations equal to each other, but ther is also the time variable?!

    Can someone please guide me. Thank you...
  2. jcsd
  3. Oct 7, 2013 #2


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    Staff: Mentor

    To gain some insight into this type of problem, try using a graph. Put time t on the horizontal axis, and put distance d on the vertical axis. Place the starting distance of the Hummer up at 20m on the distance axis, and your starting position at d=0 on the vertical axis. Both of these points are at t=0 so far, makes sense?

    Now, since the position versus time graph of the Hummer which has constant acceleration depends on t^2, what is the shape of that graph? What is it called? And your velocity is constant, so what is the shape of your d=f(t) graph? What is it called?

    And finally, the problem says that you manage to barely intercept the Hummer, so what does that mean about the two graphs crossing?

    You have equations for the f=d(t) for each of the graphs, set them equal at the one intersection point, and solve for it...

    Please show your work :smile:
  4. Oct 8, 2013 #3
    The hummer would have a parabolic shape where as the runner would be a linear function. It makes complete sense now. However, the equations have 2 variables.

    Runner: y=ax

    Hummer: y=3/2(x^2) + 20

    How would I solve that when there is an "a" variable and an "x" variable.

    I appreciate your help.
    Last edited: Oct 8, 2013
  5. Oct 8, 2013 #4


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    For the minimum a, what will be the special relationship between the line and the parabola?
    the y=ax will be tangential to the parabola, right? Write an expression for the slope of the tangent at x.
  6. Oct 8, 2013 #5
    I am not familiar with writing the tangential expression. I know that tangential (in this context) means that the runner will "intersect" the hummer once at some time t.
  7. Oct 9, 2013 #6


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    Tangential means it will touch but not cross. Varying the speed of the runner means varying the slope of the straight line. At the minimum speed, if you were to lower the slope any more then it would miss the parabola. At this point, it is a tangent to the parabola.
    Do you know how to work out the slope of a parabola at a point? Where the lines touch they willl have same slope.
  8. Oct 9, 2013 #7
    These are the equations you need. Just set y for the runner equal to y for the Hummer, and solve the resulting quadratic equation (using the quadratic formula) for the time x at which the two y's are equal (in terms of a). For any given value of the velocity a, this quadratic equation will have two solutions for x. The two solutions are real if the runner is more than fast enough to catch the Hummer, and imaginary if the runner is not fast enough to catch the Hummer. There is one case in which the two real solutions are equal, and this is the case where the runner is just barely fast enough to catch the Hummer.

    Last edited: Oct 9, 2013
  9. Oct 9, 2013 #8
    When they meet the two have the same position, y.
    Equating the two expressions you will have a quadratic equation in x (which is actually time) with a parameter a (velocity).
    You will see that for the quadratic equation to have solutions, a should satisfy some condition.
    And that condition will provide the minimum value of a (velocity).
  10. Oct 9, 2013 #9
    ax = 3/2 (x^2) + 20

    0 = 3/2 (x^2) - ax + 20

    At this point, I don't think I know how to solve for "a".
  11. Oct 9, 2013 #10
    No, you solve for x not for a.
    Do you know the quadratic formula, with that square root in it?
  12. Oct 9, 2013 #11
    (-a±√(-a)^2 -4(3/2)(20))/3

    (-a±√(a^2 - 120)) / 3
  13. Oct 9, 2013 #12
    Good. Now, in order for your quadratic equation to have only one real root, what does "a" have to be?
  14. Oct 9, 2013 #13
    Root of 120! =0
  15. Oct 9, 2013 #14
    Hu? Does this mean that [itex]a=\sqrt{120}[/itex]? If so, then you are correct. This would be the runner's speed just barely adequate for the runner to catch up with the Hummer. Anything less than this, and he wouldn't make it. Anything greater, and he would pass the back end of the Hummer before it caught up with him again.
  16. Oct 9, 2013 #15
    Yep. That's exactly what I mean. Thank you for helping me through this problem. I greatly, appreciate it... my physics teacher would freak out if I told him I needed help.

    Thank you once again.
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