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Time to decelerate over a given distance

  1. Jan 7, 2005 #1
    Alright, so the problem I'm working on is basically as such:

    An object with an initial Velocity of 95m/s slides horizontally, constantly losing velocity. After traveling 8 km it comes to a stop.

    How long did the object take to travel 8 km.

    So here's how I went at it:

    δx = v0 + .5at^2

    δx+v0 = .5(δv/t)t^2

    2(δx+v0) = v(t^2)/t

    2(δx+v0)/δv = (t^2)/t

    2(δx+v0)/δv = t

    Then I plugged in:
    δx = 8000m
    v0 = 95m/s
    δv = -95m/s

    Which returned t = -170.421s (Which I thought was odd to begin with, as how can one have a negative value for time.)

    But I tried plugging that into my initial equation to see that it all worked out but instead of getting δx = 8000m I got δx = 8190m

    So obviously there is a flaw in my logic SOMEWHERE. I just don't know where.

    Help please!?
     
    Last edited: Jan 7, 2005
  2. jcsd
  3. Jan 7, 2005 #2

    HallsofIvy

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    Science Advisor

    You appear to be using a 'special character' that shows up on my reader as ? .
    Was it δ ?

    Your first error is not "logic" but the wrong formula: δ x= v0t+ (1/2)at2. You forgot the "t" multiplying v0.
    Making that change, assuming that "constantly losing velocity" mean that the decelleration is constant, δx= v0t+ (1/2)at2. You are given that x= 8000 m and that v0= 95 m/s. That leaves a and t as unknowns. Since there are two unknowns, you need a second equation: in that same time, t, the velocity reduces from 95 to 0 so -95= at is the second equation.

    You need to solve 8000= 95t+ (1/2)at2 and -95= at.

    Substitute t= -95/a as you did and solve the quadratic equation.
     
  4. Jan 7, 2005 #3
    Ahh, so I did. Thank you.
     
    Last edited: Jan 7, 2005
  5. Jan 7, 2005 #4
    I used the formula

    Vf^2 = Vo^2 +2ad
    I found a from this. which was -0.564m/s^2

    From there I went to the formula d= Vo+1/2at^2 and solved for time.
    In which I got about 167 seconds.
     
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