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Accelerated Motion

  1. May 11, 2008 #1
    Hey I have a quick question. If the length of an inclined plane, a little ball is rolling down, measures 25 meters, and its acceleration is 2 meters every second, the velocity at one second would be 2 meters a second. The velocity at 2 seconds, would be 4 meters per second, the velocity at 3 sec is 6 meters per second. the velocity at 4 sec would be 8 meters a second, and finally at five sec the velocity would reach 10 meters per second. Now the total distance the ball travels is 25 meters. They tell you that in two seconds the ball travels 4 meters. So the ball would travel 9 meters in 3 seconds, and 16 meters in the 4 second and 25 meters in the 5th second? Adding all those up you get a number way larger than 25 meters long...what am I doing wrong?
     
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  3. May 11, 2008 #2

    dx

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    If the ball starts from rest and has a constant acceleration, the distance travelled is proportional to the square of the time since it started. So, say it travels 1 m in 1 second, then it will travel 4 meters in 2 seconds. Not 4 meters in the 2nd second. In the 2nd second, it will travel 4 - 1 = 3 meters.
     
    Last edited: May 11, 2008
  4. May 11, 2008 #3

    malty

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    What exactly is the question asking you?

    IF the total length of the plane is 25m, then clearly after travelling 25m the acceleration of 2ms^-2 will no longer be the case.So after 5 seconds the acceleration will be different. Are you given the balls initially velocity or do we assume it to be zero? Would you mind to quote the full question.
     
  5. May 11, 2008 #4
    I followed this formula: d = 1/2 * (A) * T^2

    Here is the picture given: look at attached.
     

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  6. May 11, 2008 #5
    At rest. it says!
     
  7. May 11, 2008 #6

    dx

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    zachcumer, you seem to be interpreting d as the distance travelleed in the Tth second. It is not. It is the distance travelled in T seconds.
     
  8. May 11, 2008 #7
    can you see my pic.
     
  9. May 11, 2008 #8

    malty

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    First off, that assumes initial velocity to be zero, and that formal gives the total distance travelled from 0 to whatever t may be:

    say you take you look at five seconds

    Then the d in your equation is the total distance travelled from 0 seconds through to five seconds.

    It is NOT the distance travelled during the fifth second!

    EDIT: Ahh I see dx has beat me here.
     
  10. May 11, 2008 #9
    This is exactly what my paper says: "An object starting from rest gains a speed v = at when it undergoes uniform acceleration. The distance it covers is d = 1/2 at^2. Uniform acceleration occurs for a ball rolling down an inclined plane. The plane below is titled so a ball picks up a speed of 2 meters every second; then it's acceleration a = 2 m/s^2. The positions of the ball are shown at 1 second intervals. Complete the six blank spaces for distance covered, and the four blank spaces for speeds."
     
  11. May 11, 2008 #10

    malty

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    To get the distance travelled in anyone second you find the d for t=x and t=x-1 and subtract the two:

    Example
    Your looking for the distance travelled during the 3rd second
    Fill 3 and 4 in for t into your formula d=1/2at^2
    and the difference in the two d's is the distance travelled in the 3rd second.
     
  12. May 11, 2008 #11
    v = at. so a = 2 meters per second. time lets say equals one second. In one second the ball has traveled 2 meters. lets try that in two seconds. at = 2 * 2 = 4. 4 meters in two seconds...etc..right?
     
  13. May 11, 2008 #12

    dx

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    at is the velocity. Not the distance.
     
  14. May 11, 2008 #13

    malty

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    4 metres in the second second!! In two seconds the ball has gone 6 metres!!! As dx pointed out that's a velocity [check the units]

    You told us the formula d=1/2at^2 yet you used v=at?
     
    Last edited: May 11, 2008
  15. May 11, 2008 #14
    d = 1/2 * a * t * t

    distance in 5 seconds = d = 1/2 * 2 * 5 * 5...

    Because the ball has rolled for 5 seconds.
     
  16. May 11, 2008 #15

    dx

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    Yes, thats right.
     
  17. May 11, 2008 #16
    So.....v= at ...d = 1/2 a * t * t...

    V = at.

    A = 2 meters in one second.

    T = I dunno, 2 seconds.

    2 * 2 = 4...
     
  18. May 11, 2008 #17

    dx

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    What are you trying to calculate? Is it the time taken to travel 25 meters? Also, a is 2 m/s in one second, not 2 m in one second. Remember, a is an acceleration.
     
  19. May 11, 2008 #18
    I need to find the velocity of the ball at 2 second, 3 second 4 second, and 5 second. Also need to find distances at each second.
     
  20. May 11, 2008 #19
    Look At The Picture!
     
  21. May 11, 2008 #20

    dx

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    You know that the velocity is increasing at the rate of 2 m/s per second. You also know that in the beginning, the velocity is zero. So, the velocity increases by 2 m/s every second. After 1 second it increases from 0 to 2 m/s. In the next second, it increases from 2 m/s to 4 m/s etc. You can find the distances at each second from the formula

    [tex] s = \frac{1}{2} a t^2 [/tex].
     
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