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Kinematic Equations - Why can't I solve like this?

  1. Sep 16, 2010 #1
    Ok, pretty basic Physics-101 question. However, I don't understand why my method for solving isn't working.
    1. The problem statement, all variables and given/known data
    Two cars. Car one is at rest, while the car two is traveling at a constant 16.66 m/s (given as 60 km/hr). At the moment car two passes car one, car one begins accelerating at a constant 2 m/s2. Eventually car two overtakes car one. At that point what is the distance, time, and speed of car one?

    Car one = s for sports car, car two = f for friends car.
    Here are all the relevant variables:

    car one or "s":
    vs0 = 0 = initial velocity of car one
    vs = ? = final velocity of car one
    xs0 = 0 = starting distance of car one
    xs = ? = final distance of car one
    ts0 = 0 = starting time of car one
    ts = ? = final time of car one
    as = 2 m/s2 = acceleration of car one

    car two or "f":
    vf0 = 16.66 m/s = initial velocity of car two
    vf = 16.66 m/s = final velocity of car two
    xf0 = 0 = starting distance of car two
    xf = ? = final distance of car two
    tf0 = 0 = starting time of car two
    tf = ? = final time of car two
    af = 0 m/s2 = acceleration of car two

    2. Relevant equations
    kinematic equations
    http://wiki.answers.com/Q/What_are_the_kinematic_equations" [Broken]


    3. The attempt at a solution
    So, just to get it out of the way the answers from the back of the book are: speed=33 m/s, distance=190 m, and time=11 s. I suspect these might be wrong. If something accelerates at 2 m/s2 for 11 seconds it will be going 22 m/s, not 33 m/s.

    I figured that since we are trying to find the point at which the cars meet again then the time, and distance at the end must be equal.

    ts = tf and xs = xf.

    [tex]
    x_{s} = 0 + 0t_{s0} + \frac{1}{2} * 2 \frac{m}{s^{2}} * t^{2}_{s} = 1 \frac{m}{s^{2}} * t^{2}_{s}
    [/tex]
    [tex]
    x_{f} = 0 + 16.66 \frac{m}{s} * t_{f0} + \frac{1}{2} * 0 * t^{2}_{f} = 16.66 \frac{m}{s} * t_{f}
    [/tex]
    So I can just set xs equal to xf

    [tex]
    1 \frac{m}{s^{2}} * t^{2}_{s} = 16.66 \frac{m}{s} * t_{f}
    [/tex]

    Since ts = tf, I should be able to change it to use all one or the other. Then solve for t. However, doing that gives t = 16.66 s.

    I tried the same general idea except replacing xf with the other distance formula. That ended up giving me t = 8.33 s.

    So I guess I have several questions:
    1. Why doesn't my method work?
    2. Are the provided answer correct?
    3. How do you actually solve this?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 16, 2010 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Seems right to me.

    What other distance formula?

    I think your first method does.

    No, not as far as I can tell.

    Since the car's positions are equal when one overtakes the other, you set the expressions for the positions vs. time of each car to be equal to each other.
     
  4. Sep 16, 2010 #3
    Sorry I was trying to cut down on vertical space so I just referenced the formulas in the link.
    The distance formula I used above was:
    x = x0 + v0t + .5at2
    The second distance forumla I mentioned is:
    x = x0 + .5(v0 + v)t

    The second equation uses v, which is unknown for car one, but known for car two. Since the two are both solved for x, I tried using the first one with the car one variables in it. Then set it equal to the second with the car two variables plugged in. Earlier I got 8 m/s using that method. However, just now, I realize that is because I used 0 for vf0 instead of 16 m/s.


    So I suppose the book is wrong. This is the second error in only a handful of problems we've done. I'm glad I didn't buy the book for $150. If my method is correct then I have no problems figuring out the rest. Thanks for your help.
     
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