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Car coming to a stop kinematics problem

  1. Sep 8, 2014 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    The acceleration of a particular car during braking has magnitude bt, where t is the time in seconds from the instant the car begins braking, and b = 1.5m/s3. If the car has an initial speed of 45m/s , how far does it travel before it stops?

    2. Relevant equations
    There are two equations that are relevant, I think
    [itex]\vec{v_{f}}=\vec{v_{i}} + \vec{a} t[/itex]
    and
    [itex]\vec{d}=\vec{v_{i}} t + \frac{1}{2} \vec{a} t^{2}[/itex]


    3. The attempt at a solution

    I then used the first equation to find the time it took for the car to come to a complete stop, or vf = 0.
    [itex] 0 = 45 (m/s) - (1.5 (m/s^{3}) t) (t) [/itex]
    Simplifying and solving for t I got, [itex] t = \sqrt{30} s [/itex]

    Using this in my second equation, [itex]\vec{d}=45 (m/s) \sqrt{30} s + \frac{1}{2} -(1.5 (m/s^{3})* \sqrt{30}) 30[/itex]

    Using my calculator I got 123 m/s, or 120 m/s in two significant figures like the problem asked. However, this was not right apparently. Please point my vector in the right direction!
     
  2. jcsd
  3. Sep 8, 2014 #2

    NascentOxygen

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    Staff: Mentor

    Those kinematics equations you memorised hold only for constant acceleration. So if a is varying, those equations hold only for a brief time, Δt, and you must use calculus. You will see this if you sketch the graphs, so I suggest that you first work through this problem using graphs, areas under graphs, and/or slopes of graphs. Do it all graphically. Then consider how you could perform those same operations using maths.

    Good luck!
     
  4. Sep 8, 2014 #3

    BvU

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    Your equations are applicable if a is constant, which it is not.
    You have to revert to the definitions of a and v and re-do the integration.
     
  5. Sep 8, 2014 #4
    You are given acceleration.

    The integral of acceleration with respect to time is velocity. The integral of velocity with respect to time is position.
     
  6. Sep 8, 2014 #5

    B3NR4Y

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    Okay, so acceleration is a straight, decreasing line and therefore the slope of the line is -1.5. Then integrating the equation for acceleration (-1.5 t) I got [itex]\frac{-1.5}{2} t^{2} + C = \vec{v}_{f}[/itex] and when time is zero, C is vi, so the equation for the final velocity is [itex]\frac{-1.5}{2} t^{2} + \vec{v}_{i} = \vec{v}_{f} [/itex]. And integrating that for time, I get displacement and, to spare words, I won't explain the work and I got [itex]\frac{-1.5t^{3}}{6} + \frac{\vec{v}_{i}}{2} t^{2} = \vec{d} [/itex] (I ignored putting the initial height, which I got the arbitrary constant to be because I put the origin in the problem at where the person starts braking. So solving for time I got [itex] t = 2\sqrt{15} [/itex] and using this to find distance I got ~1233.8 meters, which I think makes no sense. I don't want any more points deducted, so how else would I go about finding out if I'm wrong? My intuition tells me that braking distance >1 km makes absolutely no sense, but some of the answers on here have been more weird so I don't know.

    I actually used an internet calculator to find my answer of 1233.8 meters, and I must have made an error when I typed it in cause when I went by hand I got 232.37 m and that answer makes more sense to me, but not by much cause I don't think it should take 232.37 m to come to a stop.
     
    Last edited: Sep 8, 2014
  7. Sep 8, 2014 #6
    Where did the t^2 come from on the velocity part of your position equation? It should be a single t. Check your integration. Initial velocity does not change with time.

    Edit: you would have caught your mistake with a unit analysis. Always carry units in your calculations.
     
  8. Sep 8, 2014 #7

    B3NR4Y

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    I may have gotten a little excited when typing that part of the equation in LaTeX :P, I finally gathered all my courage and put my answer of ~232 m into the problem and it accepted it as right. Thank you everyone for your help!
     
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