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Kinematic motion problem

  1. Feb 17, 2008 #1
    [SOLVED] Kinematic motion problem

    1. The problem statement, all variables and given/known data

    A police car stopped at a set of lights has a speeder pass it at 100 km/h. If the police car can accelerate at 3.6 m/s^2,

    a) how long does it take to catch the speeder?
    b) how far would the police car have to go before it catches the speeder?
    c) what would its speed be when it caught up with the car? Is this speed reasonable?

    Answers are supposed to be a) 15s, b) 427m & c) 55m/s

    2. Relevant equations

    Not sure.

    3. The attempt at a solution

    I don't know where to start. Any help as to where to start would be appreciated.
     
  2. jcsd
  3. Feb 17, 2008 #2

    PhanthomJay

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    Please make some attempt using your knowledge of the kinematic equations of motion. Once you solve for part a), the others should readily follow. Here's a hint for part a), to get you started: when the cop catches up to the speeder, they both will have traveled the same distance from the lights in the same amount of time.
     
  4. Feb 17, 2008 #3
    what do you know about the distances travelled when the cop catches up to the speeder? The rest of the question becomes pretty easy when you remember this.
     
  5. Feb 17, 2008 #4
    Since the distance travelled by both the cop and speeder are the same,

    to find how long it took for the cop to catch the speeder,
    do you use this equation

    v2=v1 + a*t , and solve for t?

    since acceleration and velocity are given in the problem.

    I'm so used to solving problems that has all variables laid out and you choose an equation that would solve for it. If only it were that simple lol.

    I don't know which equations to use. I tried a bunch and it doesn't work.
     
  6. Feb 17, 2008 #5
    I find it is best to start off by listing all known variables.
    For example:
    V[tex]_{s}[/tex] (Velocity of speeder)= 100 km/h
    V[tex]_{pi}[/tex] (Velocity of Police Car initially)= 0 m/s
    A[tex]_{p}[/tex] (Acceleration of Police Car)= 3.6 m/s[tex]^{}2[/tex]
    (EDIT: I don't know why subscripts are showing up as superscripts... oh well.)

    You'll need to keep units consistent throughout the problem, and usually we use meters and seconds in physics problems. So convert the velocity of the speeder into m/s.
    Then write the equations that you need or might use. There are 4 or 5 kinematics equations you generally will be using for these problems with constant acceleration. You can find them here- https://www.physicsforums.com/showpost.php?p=905663&postcount=2.

    Then try to solve for the variables you don't know and use the special relationships that PhanthomJay and ||spoon|| suggested to set equations equal to each other.
     
  7. Feb 17, 2008 #6
    why not use x=ut+1/2at^2 for both the cop and speeder. Then make both equations equal to eachother since they must have the same value of x

    Edit: this will allow you to find t and hopfully solve the rest of the question.
     
  8. Feb 17, 2008 #7
    Police
    x=ut+1/2at^2
    = 1/2(3.6)t^2
    = 1.8 t^2

    Speeder
    x=ut+1/2at^2
    = (27.78)*t+1/2*a*t^2

    essentially,
    ut+1/2at^2=ut+1/2at^2
    soo basically i have to get a quadraic equation and the one of the roots will be the time?

    I'll read over all the post more carefully and try this problem once more. I'll be back when I solve it. Thanks so much for the input so thus far everyone!
     
    Last edited: Feb 17, 2008
  9. Feb 17, 2008 #8
    What do you know about the acceleration of the speeder? His velocity isn't changing. So what does that mean in terms of the accelerations?
     
  10. Feb 17, 2008 #9
    Since velocity isn't changing... the acceleration is constant? therefore a=0?

    equation for speeder would be

    Speeder
    x=ut+1/2at^2
    = (27.78)*t+1/2*a*t^2
    = 27.78*t

    since a=0 .. it would cancel out (1/2*a*t^2) ?
     
  11. Feb 17, 2008 #10
    What you should get is:

    27.78t=1/2(3.6)t^2
    (speeder x) = (cop x)

    I dont k ow where your t^2 went in the cops equation...??
     
  12. Feb 17, 2008 #11
    Oops, I edited it.
    27.78t=1.8t^2
    is what i got...

    so, to solve for t...

    1.8t^2 - 27.78t =0
    roots: t=0 and t=15.43
     
  13. Feb 17, 2008 #12
    correct. The other parts of this question should be fine now?
     
  14. Feb 17, 2008 #13
    to find c) I would do
    v=axt
    =3.6x15
    =54m/s

    to find b) I would just use one of the equation to find displacement. I think I finally got it.
     
  15. Feb 17, 2008 #14
    yep sounds good to me. :)
     
  16. Feb 17, 2008 #15
    :) Thank you very much!

    p.s I used
    x=ut+1/2at^2
    to find displacement...

    x=ut+1/2at^2
    =(0)+1/2(3.6) (15^2)
    = 405m

    the number is little off because answer should've been 427m, did I use the wrong values?
     
  17. Feb 17, 2008 #16
    no, your text has just used the rounded values fron the previous question to get 427, i assume you didnt. I got 428 aswell for the same reason. Dont worry about it ;)
     
  18. Feb 18, 2008 #17
    Off topic: You should include the variable in your tex entry, like:

    [tex]V_{pi}[/tex]

    Otherwise the difference in heights between the tex image and nearby text make it look off.
     
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