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Constant acceleration problem with a car

  1. Sep 8, 2016 #1
    1. Question to problem
    A car moving with constant acceleration covered the distance between two points 57.9 m apart in 5.02 s. Its speed as it passes the second point was 14.4 m/s. (a) What was the speed at the first point? (b) What was the acceleration? (c) At what prior distance from the first point was the car at rest?

    I'm using wileyplus.com to submit answers, and the only one that I'm stuck on is problem (c). A and B are correct.






    2. Relevant equations

    (a)V average= V final + V initial/ 2

    (b)V final= V initial + at

    (c)V final^2 = V initial^2 +2ad




    3. Attempt at solution

    (a) 11.5 m/s= 14.4 m/s + V initial / 2
    V initial= 8.60 m/s

    (b) 14.4 m/s = 8.60 m/s + a(5.02 s)
    a= 1.16 m/s^2

    (c) (8.60 m/s)^2 = (0 m/s)^2 + 2d(1.16 m/s^2)
    d=31.9 m

    For (c) it says on wiley plus that it is incorrect.
    Any suggestion on what I did wrong?





     
    Last edited by a moderator: Sep 8, 2016
  2. jcsd
  3. Sep 8, 2016 #2

    RUber

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    I am not familiar with the equation you used for (c).
    In part (a), you found the velocity at the first point. In part (b) you found the constant acceleration.
    How long did it take the car to reach the velocity from (a)?
    How much distance would a car cover from rest in that amount of time at the acceleration you found in (b)?
     
  4. Sep 8, 2016 #3

    kuruman

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    Your method is correct and if you solved the problem the way RUber suggested, you would get the same answer. I suggest that you do it more carefully, using symbols instead of numbers and plug in numbers at the very end. If you can't do that, carry your intermediate numerical calculations to more significant figures. It looks like a round off error that fell outside the tolerance of wileyplus's algorithm.
     
  5. Sep 8, 2016 #4

    jbriggs444

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    To add to Karuman's response...

    You can find that equation in various SUVAT summaries. The mnemonic I use to remember it is to mentally translate to ΔKE = work and divide out the mass.

    ##E = Fd##
    ##\frac{1}{2}mv^2 = ma d##
    ##\frac{1}{2}v^2 = ad##
     
  6. Sep 9, 2016 #5

    CWatters

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    Looks right to me. Perhaps try entering the answer using two or three decimal places?
     
  7. Sep 9, 2016 #6

    SammyS

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    I did calculations without rounding at intermediate values.

    I get a slightly larger answer for (a) and a slightly smaller answer for (b) than what you did..

    Using the square of answer (a) in the numerator and answer (b) in the denominator gives a more than slightly larger answer for (c) compared to your answer.
     
    Last edited: Sep 9, 2016
  8. Sep 9, 2016 #7

    I tried the calculation throughout without rounding for problem (c) starting with solving again for (a) and (b). I did get a slightly larger answer for (a) and a smaller answer for (b), but the were already correct beforehand when I did round them as there was a +/- 2% tolerance for answers. When I got to (c), I got 32.385... rounded it to 32.4 which was indeed correct! Just a simple rounding issue.
     
  9. Sep 9, 2016 #8

    The equation for (c) is the equation for motion with constant acceleration: V2=V02+2a(x-x0)
    where (x-x0) is the displacement in which I labeled as "d"
     
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