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Constant acceleration problem

  1. Sep 22, 2013 #1

    Kot

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    1. The problem statement, all variables and given/known data
    You are driving along a two lane high way at 50 mph. An approaching car traveling 50 mph the other way is a quarter mile ahead of you. You wish to pass a 40 ft long RV going 40 mph with its rear bumper 70 ft in front of your front bumper. Your car is 12 ft long. What constant acceleration would allow you to pass the RV and change back into your lane just as the bumpers of the RV and on-coming car are exactly your car's length apart, assuming the other vehicles remain at constant speed? What is the answer if you provide 70 ft of clearance between your car and each of the other two vehicles?


    2. Relevant equations
    ##x=x_0+v_0t+\frac{1}{2}at^2\\ v^2=v_0^2+2a(x-x_0)##

    3. The attempt at a solution
    I used the second equation to find the acceleration of my car. (50mph)^2 = 2a(122ft)
    I used 122ft for the final position of my car because my car is 70ft away from the RV, the length of my car is 12ft, and the length of the RV is 40ft which is a total of 122ft. I solved for a which is acceleration and got $54347.83 mi/h^2$ which seemed a little big. I converted it to seconds and got 15.1 mi/s^2. Is this the correct way to approach this problem?

    *edit could someone PM me how to use LaTex commands on this forum? Thanks!
     
    Last edited: Sep 22, 2013
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  3. Sep 22, 2013 #2

    Simon Bridge

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    I'll give you a quick LaTeX primer ... use the "quote" button to see the markup.

    kinematic equations:
    ##x=x_0+v_0t+\frac{1}{2}at^2\\ v^2=v_0^2+2a(x-x_0)##

    Kinematic problems are usually best solved in conjunction with a v-t diagram.
    I cannot tell from what you wrote, but to get an answer for acceleration in mi/hr2 you must make sure that all lengths are in miles and all times are in hours. Off the problem statement, that means converting that 122ft to miles.

    A useful unit for the situation would me velocity/time .. or mph/s (miles-per-hour-per-second).
     
  4. Sep 22, 2013 #3

    Simon Bridge

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    I'll give you a quick LaTeX primer ... use the "quote" button to see the markup.

    kinematic equations:
    ##x=x_0+v_0t+\frac{1}{2}at^2\\ v^2=v_0^2+2a(x-x_0)##

    Kinematic problems are usually best solved in conjunction with a v-t diagram.
    I cannot tell from what you wrote, but to get an answer for acceleration in mi/hr2 you must make sure that all lengths are in miles and all times are in hours. Off the problem statement, that means converting that 122ft to miles.

    I suspect there is something up with your calculations - cannot tell for sure without doing the problem.

    A useful unit for the situation would be velocity/time .. or mph/s (miles-per-hour-per-second).
    But it may work better to use feet and seconds for everything.
     
  5. Sep 22, 2013 #4

    haruspex

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    That will give you the acceleration needed to reach 50mph from a standing start in 122ft. You haven't used the initial quarter mile separation at all, which should be a clue that this is not right. Have another go.
     
  6. Sep 22, 2013 #5

    Kot

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    I converted all the units into SI units. I know that my initial velocity is 22.35m/s, my displacement is 36.85m (distance to change lanes when oncoming cars bumper is 3.66m from RVs bumper), I do not think I need time since this problem doesn't give any information about time, I cannot seem to find the final velocity of my car, and I am looking for the acceleration. I only know 2 of the kinematic variables but I need 3 to be able to solve an equation. Will I have to setup an equation for each car and the RV?
     
  7. Sep 22, 2013 #6

    Simon Bridge

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    Then you need another equation. You have a certain amount of time to pass the RV ... what determines that time? What haven't you used?

    Converting to SI units will risk rounding errors.
    I'd use feet and seconds, and keep fractions where they give big decimals.

    I wouldn't think about setting up equations - start with the physics, the rest will follow.
    I would do this in the frame of reference of the RV. Put x=0 on one of the RVs bumpers.
     
  8. Sep 22, 2013 #7

    Kot

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    The certain amount of time to pass the RV is when the oncoming car reaches my cars length apart from the RV? I set the origin from the bumper of my car. Should I switch it to the front bumper of the RV?
     
  9. Sep 23, 2013 #8

    Simon Bridge

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    Well, you can set any coordinates you like.

    I suspect you have the coordinate axis fixed to the road, with x=0 at the position of your front bumper at t=0.
    Your front bumper does not stay there: you are moving.

    In this system there are three things moving, one of them is going to accelerate. The only thing staying still (the road) is something you don't care about. Using this one, the distance your front bumper has to go is the distance between you and the RV, plus the length of the RV, plus the length of your car, plus the distance the RV travels in all that time.

    This means you need 3 position-time equations, and two equations describing the conditions.

    Headache much?

    If you fix the coordinate system to the RV, then it turns the problem of having to overtake a moving vehicle with the simpler one of having to overtake a stationary vehicle. On top of which, the relative speeds turn out to be nice(ish) numbers in ft/s ;)

    Don't get me wrong, there is no "right" way to do this.
    Pick one and we'll use it OK? Play to your strengths.
     
  10. Sep 23, 2013 #9

    haruspex

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    Yes
    It doesn't matter where you set the origin as long as you are consistent.
     
  11. Sep 23, 2013 #10

    Kot

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    So I would have to take into account the amount of time and make sure that my acceleration would allow my car to pass the RV before that time?
     
  12. Sep 23, 2013 #11

    Simon Bridge

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    ... but it does matter which reference frame you pick.

    ... yep, you want to keep the coordinate axis fixed to the road?

    write out the postion-time equations for each moving object.
     
  13. Sep 23, 2013 #12

    Kot

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    I'm a little confused as to why it matters which reference frame I choose, could you please explain? Thanks.
     
  14. Sep 23, 2013 #13

    Kot

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    *edit sorry double post
     
    Last edited: Sep 23, 2013
  15. Sep 23, 2013 #14

    Simon Bridge

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    Changes the speeds you use... depending on where the axis is attached.

    i.e. in the RV reference frame, the RV is stationary, you are doing +10mph and the oncoming car is doing -90mph.
    But those speeds are the same whether you put the origin on the front or rear bumpers of the RV, or 12ft ahead of the RV.
     
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