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Homework Help: Calculating minimum amount of time

  1. Sep 28, 2005 #1
    If an object starts from a point a certain an amount of time ahead of the other ,how to calculate the minimum amount of time difference for which it is impossible for object 2 to catch up on object 1? This the information;

    Object 1 starts at a initial velocity of 0 with a constant acceleration of 1m/s^2.
    Object 2 starts at a initial velocity of 4.5m/s with a constant acceleration of 0.

    Let the amount of time be u.

    Calculating the discplacement of object 1 after u.


    Calculating the final velocity;


    Now the discplacement of object 2:


    The discplacement of object 1:

    d=t^2/2 + ut + (u^2)/2

    Now for what value of u,

    t^2/2 + ut + (u^2)/2 > 4.5t

    for any value of t?
  2. jcsd
  3. Sep 28, 2005 #2


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    I believe this is incorrect. Go back to the fundamental equation of x(t).
  4. Sep 29, 2005 #3


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    Hmmm, this is an interesting one. Don't know what methodology you are suppose to use, but I can think of 2 valid approaches. By the way, it always helps to graph things if you can (Excel is a great tool for this). So:

    1. what do you know?

    [itex]x_1 = f_1(t)[/itex]
    [itex]x_2 = f_2(t)[/itex]

    2. what do you want?

    How much 'head start' in time 1 object needs so the other doesn't catch up

    3. Use what you have to get what you want

    First, you have to decide which one needs a head start. Since one is a linear function of t and the other a quadratic function of t, intuition says the linear function is the one that needs a head start (because its distance increases at a much slower rate than the other one)

    So, how to express the 'head' start in mathematical terms using what you have? Calling [itex]t_0[/itex] the head start time, aren't the distances traveled really the following:

    [tex]x_1(t) = f_1(t+t_0)[/itex] ....need to think. Is this the one w/ head start, or the other one?
    [tex]x_2(t) = f_2(t)[/itex]

    You are looking for the right [tex]t_0[/itex] such that the distances traveled by the 2 objects are the same. Hope this helps get you going in the right direction. Hint: solving a quadratic equation is involved. Actually, this method not only tells you what [tex]t_0[/tex] is, you also get the time at which the objects meet.

    The second approach is actually much easier, but not terribly intuitive. Graphing helps. The solution has to do with slopes.

    P.S. This is high school?
    Last edited: Sep 29, 2005
  5. Sep 29, 2005 #4


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    You're on the right track, but it's a little easier than that.

    If they both started from the same point, the object travelling 4.5 m/s would initially start out ahead. Eventually the accelerating object will catch up and continue to get further and further ahead. The accelerating object would catch up to the 4.5 m/s object eventually no how big of a head start the 4.5 m/s object had.

    The distance between the two objects is going to be [tex]\frac{1}{2}t^2 - 4.5 t[/tex]
    If you could find the maximum distance between the two objects before the accelerating object begins to catch back up, you'd know the minimum head start the accelerating object needed.

    How do you find the maximum distance between the two objects? It's at the point where the derivative of the distance of the two objects is zero.
  6. Sep 29, 2005 #5


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    Good one - I guess I always pick the hard way. OK, there are at least 3 valid approaches, and I've described the 2 hardest.
  7. Sep 30, 2005 #6


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    Your second way is the same as what I posted. It is the easiest way.

    Of course, if he's in high school, he might not have learned derivatives, in which case he'll have to settle for the finding the two points where the distance is zero. Since the scenario is described by a quadratic equation (which describes a parabola), it would be a pretty good bet that the maximum distance is going to occur half way between the zero points.
  8. Sep 30, 2005 #7


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    I now understand why what I did at first (never posted it) didn't work, which was to just set the 2 equations equal the each other and solve for t. The reason is that this just represents the situation where both leave at the same time - not relevant to the problem except that it confirms that a delay is needed.....
  9. Oct 1, 2005 #8
    Thank you all but I solved it...

    No need for calculus

    This is how I proceeded...

    Let x be the amount of time between each object departure.

    The time that object 1 has is t.

    The time that object 2 has i t-x

    If the two objects catch up;

    a(t^2)/2 = 4.5(t-x)

    t^2/2 = 4.5(t-x)

    * -1 ; -t^2/2 = 4.5(x - t)
    x - t = -t^2/9
    x = -t^2/9 + t
    = -1/9*(t^2 - 9t)
    = -1/9*(t^2 - 9t + 4.5^2 - 4.5^2)
    = -1/9*((t - 4.5)^2 - 4.5^2
    = -1/9*(t - 4.5)^2 + 2.25

    The time x depends on the time t. The maximum value of x is 2.25s

    So the minimum amount of time that can ellapse between the two objects departure before it's impossible for object 2 to catch up object 1 is

    2.25 + y

    where y->0
    Last edited: Oct 1, 2005
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