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Magnitude of deceleration

  1. Feb 4, 2005 #1
    An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When the engineer first sees teh car, the locomotive is 210 m from the crossing and its speed is 10 m/s. If the engineer's reaction time is .22 s, what should be the magnitude of the minmum deceleration to avoid an accident? Answer in units of m/s^2
  2. jcsd
  3. Feb 4, 2005 #2
    First, use the formula d=vt to find the distance he travels before applying the brakes. Subtract that distance from 210 m. Then, use the formula:
    [tex]d=V_it + \frac{1}{2}at^2[/tex]
    to find the acceleration, which would be a negative number.
  4. Feb 4, 2005 #3
    wait but shouldnt i solve for time 210=10(t) then t is 21 then 21-.22=20.78 then 210=10(20.78)+1/2(a)(20.78)^2 so a = .0101897345
  5. Feb 4, 2005 #4
    No, since his speed is not 10 m/s throughout. First, find the DISTANCE he travels in 0.22s (d=(10 m/s)(0.22s)).
  6. Feb 4, 2005 #5
    210-2.2=d=207.8=10(22)+1/2(a)(22)^2 ? ? or is 21.78 for time?
  7. Feb 4, 2005 #6
    You got the first part (the 207.8m) right. As for the rest, you canot use 22s and the time, since you don't know the time it will take hime to stop. (Sorry, I gave you the wrong formula).
    Ok, here's what you know:
    initial velocity=10 m/s
    final velocity = 0 m/s (since he has to stop)
    d = 207.8 m

    Can you find a formula that fits?
  8. Feb 4, 2005 #7


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    Where are you getting t=22?

    I wouldn't use the above equation.

    There's a different distance equation you can use and solve for "a" immediately... hint: it doesn't have a t in it.
  9. Feb 5, 2005 #8
    how is time 17.21 seconds i have 207.8=10t-1/2(.2406)t^2
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