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Physics kinematics SIN question

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A car, travelling at a constant speed of 30m/s along a straight road, passes a police car parked at the side of the road. At the instant the speeding car passes the police car, the police car starts to accelerate in the same direction as the speeding car. What is the speed of the police car at the instant is overtakes the other car?

    Given: v=30m/s
    vi=0

    Need: vf=?

    2. Relevant equations

    vf=vi+αΔt
    vf2=vi2+2αΔd
    v=Δd/Δt


    3. The attempt at a solution

    So far, I really have not gotten anywhere. I believe what I have to do is somehow manipulate the velocity equation of the first car into something I can input into the vf equation for the police car. I have been having much trouble with this question and would appreciate any tips to point me in the right direction.
     
  2. jcsd
  3. Dec 30, 2011 #2
    This isn't solvable without knowing the acceleration of the police car, without it the velocity when the police car overtakes the other car could be anything.

    edit: You don't necessarily need the acceleration, but you need at least one other piece of information (such as at what distance did the police car overtake the other car) to solve the problem.
     
  4. Dec 30, 2011 #3
    Yes, and that is what I have been fretting over this whole time. They give multiple choice answers, but essentially they all work. I know that depending on the magnitude of the displacement or the time, the rate of acceleration will change.
     
  5. Dec 30, 2011 #4

    berkeman

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    Interesting. I think I was able to solve it just with the given information (unless I did something wrong). Pretty simple answer too.

    You should write an equation that equates the distance travelled to the meeting/passing spot for each car (call that distance D). The speeding car's velocity is constant, so what is the equation for the time it takes for the speeding car to get to D?

    And what is the equation for the police car's time to get to the distance D? If you set those two equations for D equal to each other...
     
  6. Dec 30, 2011 #5

    SammyS

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    Yes, berkeman is correct.

    The question only asks for the speed of the police car at the moment it overtakes the speeding car. It doesn't ask for the time or distance, either of which would require more information.
     
  7. Dec 30, 2011 #6

    NascentOxygen

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    It's a trick question. They should have printed a smiley near it. So now I'm curious to see what are the answers they give you to choose from?
     
  8. Dec 30, 2011 #7
    I disagree, I still think there are an infinite amount of answers, like NascentOxygen said. Berkeman is right, except that the distance D could be anything, and the acceleration, and thereby the velocity of the police car depends on that D, so without it you can't give an answer. Or more correctly, any number is an answer.
     
  9. Dec 30, 2011 #8

    berkeman

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    Just try the math that I outlined. Don't post it for the OP to see, but you should be able to see that the question has one answer. :smile:
     
  10. Dec 30, 2011 #9
    You're completely right, my common sense distracted me from the idea that the math might all work itself out to be equal, I got the answer and it makes sense.
     
  11. Dec 30, 2011 #10
    Alright, I tried something out and I am not entirely sure if it even makes sense.

    For the speeding car:

    Δd=vΔt
    Δt=Δd/v

    For the police car:

    Δd=vfΔt-0.5aΔt2
    Δt=√Δd/0.5a

    Next I did the math:

    vΔt=vfΔt-0.5aΔt2
    30(Δd/30)=vf(√Δd/0.5a)-0.5a(√Δd/0.5a)2
    60Δd=vf(√Δd/0.5a)-(√Δd2/0.5a)
    60=vf

    the multiple choice answers are: a)45 b)30 c)60 d)75 e)100
     
  12. Dec 30, 2011 #11

    SammyS

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    Yes, that's the correct answer.

    Notice that in your equation, vΔt=vfΔt-0.5aΔt2,
    you could divide by Δt to simplify things before substituting.

    The really quick way to get the answer is:
    Using average velocity, vAvg = Δd/Δt .

    Both cars travel the same distance in the same time. Therefore, they must have the same average velocity.

    The average velocity of the speeding car is 30 m/s .

    For the police car, which has uniform acceleration, [itex]\displaystyle v_{Avg.}=\frac{v_F+v_I}{2}\,.[/itex] The initial velocity is zero, so the final velocity must be 60 m/s, since the average velocity is 30 m/s.
     
  13. Dec 30, 2011 #12
    Thanks for the help everyone, I will keep this in mind next time I am answering these kinds of questions! :)
     
  14. Dec 31, 2011 #13

    NascentOxygen

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    Any hint that we should assume a uniform acceleration is so well hidden that I still can't see it.
     
  15. Dec 31, 2011 #14
    I think that is just assumed, if it wasn't uniform there would be an infinite number of possible answers, this seemed like a basic physics 1 question so I assumed a constant acceleration.
     
  16. Dec 31, 2011 #15

    NascentOxygen

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    This is the sort of question where one teacher may justify an expectation that uniform acceleration will be assumed, and give marks for that. But at the same time another can put the case that no such assumption should be made since it was not stated, and proceed to deduct marks if it was assumed. It's a good trick question, and can serve both philosophies.

    Few vehicles could be assumed able to accelerate uniformly from 0 to 216km/hr!

    My answer remains [itex]speed \gt 30 m/sec[/itex] :smile:
     
    Last edited: Dec 31, 2011
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