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What is the speed of the cart 3.5 s after the fan is on?

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  1. Nov 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A 2.34-kg cart on a long, level, low-friction track is heading for a small electric fan at 0.22 m/s . The fan, which was initially off, is turned on. As the fan speeds up, the magnitude of the force it exerts on the cart is given by at2, where a = 0.0200 N/s2. What is the speed of the cart 3.5 s after the fan is turned on?

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    a=F/m
    =0.02t^2/2.34 *t^2
    =0.0085t^2

    v=0.205(3.5)^3+ 0.22
    =9.00m/s
    what is my error?
     
  2. jcsd
  3. Nov 2, 2016 #2

    TSny

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    How did you arrive at the number 0.205 in the expression for v?

    What is the direction of the acceleration compared to the direction of the velocity?
     
  4. Nov 2, 2016 #3
    So your calculation for the first part is mostly fine, but, the cart will be decelerating as it heads towards the fan so this acceleration value must be negative.
    It really helps if you draw a diagram of a physics problem so you can keep track of what is where and in which direction.
    The second equation: v = at + vo is indeed the right one to use, but I can't figure out where this 0.205 has come from... you just calculated the acceleration, it should read something like this:
    v = (-0.0085)t2(t) + vo
     
    Last edited: Nov 2, 2016
  5. Nov 2, 2016 #4

    TSny

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    Hello Garth Vader. Welcome to PF!

    When giving help on homework, we try to give "minimal" help to get the student on track. Then the student can still do as much of the work as possible. We definitely try to avoid showing the complete solution. (I don't think your expression for v is correct.)

    Of course, there is a gray area here, with different opinions about how much help is too much. I know I've been guilty of giving away too much information at times.

    Here's where you can read the rules for PhysicsForums.
    https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/
    There's a section on homework guidelines.

    Thanks for being willing to help out here at PF. Hope you stay around.
     
  6. Nov 2, 2016 #5
    Ah yes, will try not give too much away next time.
     
  7. Nov 2, 2016 #6
    acceleration is in the opposite direction so there should be a negative sign in front. i got the number for speed by multiplying by two during intergation
     
  8. Nov 2, 2016 #7

    TSny

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    Yes, the acceleration is negative. Check the integration. The factor of .205 is incorrect.
     
  9. Nov 2, 2016 #8
    i got the equation for v now as v= -0.00283t^3+ 0.22
     
  10. Nov 2, 2016 #9
    getting an answer of 0.098
     
  11. Nov 2, 2016 #10

    TSny

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    That looks correct. Of course you should include the proper units with your answer.
     
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