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Equations for Accelerated Motion 4

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A motorcycle traveling at 30 m/s comes to a stop over a distance of 150 m. What was the motorcycle's acceleration?


    2. Relevant equations
    v^2=vinitial+2aΔx


    3. The attempt at a solution
    I tried plugging in what I know.

    0=900+2a*150 m

    Then I divided 150 m and 900 by 150:

    0=6+2a

    This yields an answer of -4. It doesn't seem right, but I can't place why. I think I may have needed to divide the v^2 by 150, but that's 0/150, which wouldn't make sense. Thanks in advance for a reply!
     
  2. jcsd
  3. Oct 20, 2013 #2
    does it?....check your final equation, you have done the rest of it correctly....it should be -ve because of deceleration (retardation), it just shows acceleration acts in direction opposite to its initial velocity

    ##0=6+2a##
    ##a=-3##

    Edit: I think it was a typo, but this is what the equation is v^2=(vinitial)^2+2aΔx.....your calculations are correct except the very end of it.!!!
     
    Last edited: Oct 20, 2013
  4. Oct 20, 2013 #3
    Oh! Okay. I made a mistake in the post, I actually thought it was -2 m/s . I see what you did there, I added in my head instead of dividing. The main reason I thought it was wrong didn't have a reason. It just seemed like it should be different because I couldn't place how an acceleration of -2 or -3 would be enough to slow the car, but now that I look again, it makes sense. Thank you for a polite and helpful reply!
     
  5. Oct 20, 2013 #4
    The car?? I meant the motorcycle. Sorry. Blame it on lack of coffee syndrome.
     
  6. Oct 20, 2013 #5
    no problem!!!.........just to make it seem reasonable you can use ##v=v_{0}+a.t##, solving for time gives ##10~seconds##. It doesn't stop immediately, it slowly decelerates therefore time is comparatively large and so is the distance. I hope this makes more sense now!!!!
     
  7. Oct 20, 2013 #6
    I got that, no problem.
     
  8. Oct 20, 2013 #7
    It does! Thanks, you have given some of the nicest replies I have found on here. If you have a moment, maybe you would go to my other Equations for Accelerated Motion problems and check those?

    Thanks again!!
     
  9. Oct 20, 2013 #8
    HA! :rofl:
     
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