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Solving for the minimum speed a car must have to catch up to another car?

  1. Sep 13, 2012 #1
    1. A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.9 m/s2. A green car arrives at the position of the stop-light 4 s after the light had turned green. What is the slowest constant speed which the green car can maintain and still catch up to the blue car?
    Answer in units of m/s

    So, we're given that ##a = 0.9 m/(s^2)##, ##v_0 = 0##, and that displacement at t = 0 is 0 (all of this for the blue car, or ##C_B##)
    For the green car(or ##C_G##), we're given that the car is holding a constant speed, so there must be no acceleration. If there's no acceleration, then the final and initial velocities must be equal for ##C_G##. Additionally, we're given that its displacement = 0 at t = 4 (since it started 4 seconds late).



    2. We know that the velocity for ##C_B## is $$v=0.9t$$ We know that the displacement is equal to $$0.45t^2$$
    For ##C_G##, we know that ##v## is a constant c, that acceleration = 0, and that the displacement is represented by the function ##vt - 4v## (I believe).




    3. I figured that we were looking for the minimum speed that ##C_G## would have to possess to cause the displacement of both ##C_G## and ##C_B## to be equal, because this would represent that ##C_G## had caught up to ##C_B##. However, I kept getting it down to things like $$0.45t^2 = vt - 4v$$ which you can't solve to my knowledge. So I tried a graphical method. I know that the displacement of ##C_B## is a parabola, and that the displacement of ##C_G## is a linear equation that passes through the point (4,0) [if we graph displacement as y and time as x). So we're looking for a tangent that passes through the point (4,0). However, we don't know the time or displacement this occurs at.

    Sorry if this is all too verbose and such, first time posting here so I apologize!
     
    Last edited: Sep 13, 2012
  2. jcsd
  3. Sep 13, 2012 #2

    Simon Bridge

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    You need the other simultaneous equation. :D

    You forgot that this is a minimization problem ... you have an equation with two unknowns: that means there are are a range of possible values that v can take for different times to catch up. You want to find the minimum value out of this range ;)
     
    Last edited: Sep 13, 2012
  4. Sep 13, 2012 #3
    Hi Pasta - as a fellow new guy - you're going to LOVE PF!!!
     
  5. Sep 14, 2012 #4


    For real root,
    b2>4ac

    v2>4(0.45)4v
     
  6. Sep 14, 2012 #5
    Thank you Simon! When you refer to the other simultaneous equation, what are you referring to? I could only find one equation that relates them. I was thinking of using ##v_B = 0.9t##, but I didn't think that would work because ##v_G## and ##v_B## are different, so I'd still be left with two variables.

    And aziz, what do you mean? I see what you're doing, but I don't follow where to go from there. In its most basic form it seems to be saying that ##(v_G)^2 > v_G##, which only tells me that ##v_G > 1##. Or am I misreading your post? Sorry!

    And thank you enosis! I hope so. :]
     
  7. Sep 14, 2012 #6
    v>7.2m/s
    So the minimum velocity is 7.2m/s.
    Less than 7.2m/s the green car couldn't catch the blue car.

    There is no problem if the green car's velocity higher than 7.2m/s
     
  8. Sep 14, 2012 #7

    Simon Bridge

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    azizwizi is giving you the method from understanding quadratics ... I'll leave that to him.
    He's basically saying that you have a relation between v and t in which not all values of v are possible ... so you can investigate that relation to find out what the smallest possible value of v is :)
    You have not used one of your constraints - namely, that the speed has to be a minimum.

    You found v(t) ... that's equation 1.
    The variable t in this equation was the amount of time it takes to catch up.

    v(t) is a minimum for catch-up time t when dv/dt=0. ... that's equation 2.

    for smaller catch-up times, v is bigger, for longer catchup times v is smaller
    ... though, there will be some v with no corresponding t because you are going too slow to ever catch up.

    Imagine: if you are in the green car, and you are going fast, then you will catch up to the blue car at time ##t_1## ... then overtake it, but the blue car is always accelerating ... so it will start gaining on you and overtake at time ##t_2##. Therefore there are two times in which you and he are level with each other. Each time, you and he draw level, you will be doing different speeds ##|v_b(t_2)-v_g|=|v_b(t_1)-v_g| > 0##.

    If you go faster ... then those times will get further apart. The difference in your speeds when you draw level will also be greater.

    If you go slower, however, the two times get closer together. The difference in your speeds also gets smaller. It should be possible to find a speed where the two times are the same ##t_2=t_1=t## and, therefore, the difference in speeds when you catch up is zero! ##v_g(t)=v_b=v##

    If you go slower than this - the blue car will never catch the green car.

    So now you know what time the cars meet up for the slowest pursuit-speed that is capable of this you can do the rest.

    If you draw the v-t diagrams, on the same axis, for this situation - you will quickly see that the time they meet has to be t=2x4=8s: do it and see!
     
  9. Sep 14, 2012 #8
    I see how both you guys came to your methods/answers now. Thank you! Made this so much easier. :]
     
  10. Sep 14, 2012 #9

    Simon Bridge

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    No worries.
    These more open questions are closer to what you end up doing in real life ... the fun part is that there are usually many ways of thinking about them.
    The take-away lesson here would be that you don't always get just one answer ... you have to find some way to select between the ones that you have.
     
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