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Grade 11 Motion Question

  1. Feb 19, 2017 #1
    I have solved the question below and was wondering someone can have a look at it and give some feedback, please.

    1. The problem statement, all variables and given/known data
    1. An object is pushed from rest across a sheet of ice, accelerating at 5.0 m/s2 over a distance of 80.0 cm. The object then slides with a constant speed for 4.0 s until it reaches a rough section which causes it to stop in 2.5 s.
    a. What is the speed of the object when it reaches the rough section?
    b. At what rate does the object slow down once it reaches the rough section?
    c. What total distance does the object slide throughout its entire trip?

    2. Relevant equations

    vf = vi+a.t
    d = vi.t + 1/2 a.t^2
    vf2 = vi2 +2.a.d



    3. The attempt at a solution

    For part a using the given accelration and distance I found the velocity to be 2.8 m/s

    However, for part b and c I am confused and although I solved to get the answer but I feel my answers are not correct.

    b. So when the object reaches the rough section its velocity is still 2.8 m/s, and it stops after 2.5 s So using the first equation above I got the accelration to be -1.12 m/s2 by dividing 2.8 with 2.5s.

    c. Fort this last question I divided the above scenario into three parts, first part the distance is 0.8m (its given). For the second part only time is give, which is 4.0 s but I know the velocity from question a, so using the time and velocity I figured out the acceleration which was 0.71 and then using all this information I found the distance of the second part to be 16.96 m, which seems high to me. Finally, to find the distance for the last part, only time is given, 2.5s. But from question b I know that the accelration is -1.12. So then I found the velocity with this information and finally using the third equation above I found the distance for this part to be 3.5m.

    Adding all three parts I got the total distance of 21.26m.

     
  2. jcsd
  3. Feb 19, 2017 #2

    kuruman

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    Hi aatari and welcome to PF.

    :welcome:

    Parts (a) and (b) are correct. Over the middle interval of 4.0 s the speed is constant, i.e. there is no acceleration. How much distance does the object cover in 4.0 s if its speed is 2.8 m/s?
     
  4. Feb 19, 2017 #3
    I see what you mean. So if that's the case, meaning the speed is constant and there is no acceleration, do we then say that the accelration is 0? Because if we do that then we are left with speed and time and if we multiply we get 11.2.
     
  5. Feb 19, 2017 #4

    kuruman

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    That is correct. You can think of acceleration as the "speed of the speed". An object's position stays the same when its speed is zero. Likewise, an object's velocity stays the same when its acceleration is zero.
     
  6. Feb 19, 2017 #5
    So for the third part where the object reaches the rough surface and stops after 2.5 seconds. Is it correct then that I have the distance of 3.5 m and if I combine all the distances, I end up with 15.5m.
     
  7. Feb 19, 2017 #6

    kuruman

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    That would be correct.
     
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