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Simple physics question

  1. Aug 29, 2005 #1
    Is it true that if the rate of decceleration is two times slower than the rate of acceleration for the same object, that it will need two times the distance to come to a stop as it did to accelerate?
     
    Last edited: Aug 29, 2005
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  3. Aug 29, 2005 #2

    russ_watters

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    The answer to this question depends on if this is a homework question... :biggrin:
     
  4. Aug 29, 2005 #3
    It is but it is for extra credit :rofl:
     
  5. Aug 29, 2005 #4

    LeonhardEuler

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    It's not a very dificult question. Find the distance travelled under constant acceleration for a given period of time. Use what you know about how much the velocity change to solve for the time taken in terms of the maximum velocity and acceleration. Plug this in to the equation you had for the distance travelled. Now you have the distance traveled in terms of the acceleration and maximum velocity. By seeing how the acceleration varies with distance your problem is solved.
     
  6. Aug 29, 2005 #5

    Doc Al

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    Figure it out for yourself! How does the distance required to accelerate from a speed of 0 to V (or decelerate from V to 0) depend on the acceleration? (Work out or look up the kinematic formulas for uniformly accelerated motion.)
     
  7. Aug 29, 2005 #6
    I have no idea where to start or how the distance is related to the acceleration.
     
  8. Aug 29, 2005 #7

    Doc Al

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    Are you familiar with the kinematics of uniformly accelerated motion? (If not, then you are not ready to solve this problem.)
     
  9. Aug 29, 2005 #8
    I know that the equation for acceleration is final velocity - initial velocity divided by time, but I really don't see how that helps answer my original question :(
     
  10. Aug 29, 2005 #9

    Doc Al

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    That's a start: You have a relation connecting acceleration and velocity (and time). Now figure out how to get the distance. (Hint: Distance equals average speed X time.)
     
  11. Aug 29, 2005 #10
    The problem doesn't give me average speed or time though.
     
  12. Aug 29, 2005 #11

    Doc Al

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    You don't need actual values. You are trying to find the relationship between acceleration and distance. So far you have:
    (1) a = v/t (the initial speed is zero)

    I suggest you combine that with my "hint" about distance and average speed. (If something goes from 0 to V, what is its average speed?) Write my statement mathematically. Then you'll have two equations that you can combine to relate distance with acceleration.
     
  13. Aug 29, 2005 #12
    So equation 2 would be x = v*t and you substitute v = at and get

    x = a*t^2

    Is that right?

    Wow I feel stupid thanks for the help on that. :smile:
     
    Last edited: Aug 29, 2005
  14. Aug 29, 2005 #13

    Doc Al

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    No.... you need average speed, which is v/2, not v.
    When you combine the equations, don't eliminate v; instead, eliminate t. You want to end up with an equation relating v, a, and x.
     
  15. Aug 30, 2005 #14
    So instead equation 2 would be

    x=(vt)/2

    Solving for t would be

    t = 2x/v

    Substituting that into equation 2 would be:

    a= v * (v/2x) or a = v^2 / 2x

    Meaning when acceleration is 2 times decceleration then you need half the distance to accelerate? Thanks I think I get it now, I am not worthy :P
     
  16. Aug 30, 2005 #15

    Doc Al

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    Exactly. The derived relationship a = v^2 / 2x tells you that acceleration and stopping distance are inversely related (for a given maximum speed).

    Note that this same equation can help you understand why, for a given deceleration, that if you are moving twice as fast it takes 4 times the distance to stop.
     
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