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Supposedly Simple, 1-D Motion Question

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A man is driving at 56.0 km/h [N]. Suddenly, a boy stumbles on the road 65 m ahead of the car. After t seconds the driver finally decides to hit the brakes, which then produces an acceleration of 3.00 m/s² . What is the maximum reaction time allowed if the driver is to avoid hitting this student?

    2. Relevant equations
    I started off trying to use Δd = V1Δt + 1/2a(Δt)² , however I was told this won't work since acceleration is not constant, this I understand. Then, I tried using quadratic formula to solve, but for me it wouldn't work since the number I was trying to find the square root of was negative.

    3. The attempt at a solution
    I would write the steps I took during my quadratic formula phase, but I'm not sure I even did it correctly, any help with this would be appreciated.
     
  2. jcsd
  3. Sep 22, 2008 #2

    LowlyPion

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    Just write out your quadratic. You don't need to show the intermediate steps.
     
  4. Sep 22, 2008 #3
    0 = (-1.5m/s²)(Δt)² + (16m/s)(Δt) - 65m
    X1= 5.3 X2 = 5.3

    That's not the answer I'm supposed to be getting.. What have I done wrong?
     
  5. Sep 22, 2008 #4

    LowlyPion

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    OK I get 15.55m/s instead of 16 for your equation.

    But perhaps a different approach might be easier?

    How long does it take to slow from 15.55m/s to 0 at -3m/s2? That's your time budget to stop.

    Using that time, determine how much distance you require to stop.

    If it is less than 65m then then how long can you continue at 15.55m/s before you have to stop?

    If it is longer than 65m, start calling 9-1-1.
     
  6. Sep 22, 2008 #5
    0 m/s = 15.55 m/s + (-3.00m/s²)(Δt)
    -15.55m/s/-3.00m/s=(Δt)
    (Δt)=5.183
    Δd=15.55m/s(5.183s)+1/2(-3.00m/s²)(5.183s)²
    Δd=40.3m
    65m - 40.3m = 24.7m/15.55m/s= 1.58 s?

    This is about two decimal places off from the given answer (1.56) however, is the procedure correct?
     
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