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Rolling Ball and Proportionality Between Distance and Time

  1. Sep 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A ball starts from rest and rolls down a hill with uniform acceleration, traveling 130m render?expr=%7B%5Crm+m%7D.gif during the second 6.0s render?expr=%7B%5Crm+s%7D.gif of its motion.
    How far did it roll during the first 6.0s render?expr=%7B%5Crm+s%7D.gif of motion?


    2. Relevant equations
    I guess:

    v=v(0)+at
    v^2=v(0)^2+2ay
    x=x(0)+v(0)t+0.5at^2


    3. The attempt at a solution
    O
    kay, if I am not mistaken, then the distance travelled by the ball is proportional to the square of the time passed.
    d is proportional to t^2
    I have tried so many answers already.
    I have tried setting the second six seconds to be twice the time and then the distance is one fourth of 130m. That was wrong. I tried setting the time to twelve seconds. I have tried dividing the distance by two. If I set the time to be six seconds and the distance to be 130 meters, then I don't have the initial velocity. If I set the initial velocity to equal zero and then the time to be six seconds, I don't have the distance. I tried making the time proportional to the square root of the distance. I only have one more attempt, and this question threw me off because it gave me info about the second six seconds and not the first six seconds. If it were to ask me "How far did it go in twice this time?" then I would be able to answer it.

    But I really cannot find this answer. I have tried 35 meters, 11 meters, 130 meters, 98 meters, 22 meters, 65 meters. Nope. I figure it must be less than 130 meters because the ball is accelerating down the incline and it would go faster every second because the velocity is increasing. Please help me. Please.
     
  2. jcsd
  3. Sep 21, 2014 #2
    the ball traveled 43.33 meters....
    first of all, you use the equation delta x = v initial * time + 1/2 acceleration * t^2, time is 6 and delta x is 130 meter.
    you get 130 = 6 v + 18 a

    secondly, you use the formula acceleration = v initial - v final, NOTICE this equation is for the first 6 seconds, so the v initial starts off as zero and v final is the v initial in the first equation. plug in v initial is zero, time is 6.
    you get a = v / 6

    thirdly, you combine both equations by subbing in "a" from the second equation into the first equation.
    you get 130 = 6 v + 18 (v/6)
    you simply. V = 130 / 9

    now you have the final velocity for the first 6 seconds. you use the equation v^2=v(0)^2+2ad where V is 130 / 9, v0 is zero since it starts from rest and a is v/6 which is 130 / 9 / 6.

    you plug the number into the equation you get
    (130/9)^2 = (0)^2 + 2 * (130 / 9 /6)*d
    and you go the calculations to solve for distance which in my case is 43.33 if my calculator is not broken.
     
  4. Sep 21, 2014 #3
    ops,, in the second step the formula is acceleration = (v initial - v final) / time
     
  5. Sep 21, 2014 #4
    Thanks a lot. I didn't know it was so complicated. Thanks again.
     
  6. Sep 21, 2014 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It is much easier.

    Let's call the distance it travels for the first 6 seconds x. As distance goes with time squared, after twice this time (so after 12 seconds) it traveled 4x. The difference between those two values is 3x, and we know it is equal to 130m...
     
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