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Solving this motion problem

  1. Sep 29, 2012 #1
    1.There are two parts to this question, and I can get the first part but do not know where to start for the second.

    It goes like this; You are driving down the highway late one night at 20m/s when a deer steps out on the road 35m in front of you. Your reaction time before stepping on the brakes is 0.50s, and the maximum deceleration of your car is 10m/s2.

    a) How much distance is between you and the deer when you come to a stop?
    b) What is the maximum speed you could have and still not hit the deer?


    I did part a by finding the distance of 20m/s(0.50s) = 10m then adding it to the distance calculated from V22-V12 / 2(a) = 20m
    so it came out to be 30m and 5m away from the deer.

    How do I calculate the max speed?

    Thanks for any help
     
  2. jcsd
  3. Sep 29, 2012 #2
    Try using the same equation you used to solve part A, only solve for the initial velocity this time.

    V2= 0 m/s
    deltaX = 35m (or maybe 34m so you wouldn't technically touch the deer?)
    a = -10 m/s^2

    That doesn't really take into account the 0.5sec to brake, but I don't know how to account for that.
     
  4. Sep 30, 2012 #3
    Yeah I tried that, it doesn't work out.

    The answer is supposed to work out to 22 m/s.
     
  5. Sep 30, 2012 #4
    (b)
    Just draw a velocity vs. time graph.
    Area under the horizontal line joint to a line with gradient of -10 is equal to 35.
     
    Last edited: Sep 30, 2012
  6. Sep 30, 2012 #5

    ehild

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    You have to take the 0.5 s into account. So the distance along you can decelerate is 35-0.5vmax. The time available for deceleration is vmax/10.

    ehild
     
  7. Sep 30, 2012 #6
    Okay guys, I solved for part b and ended up with the correct answer. Im not confident in my method as it took me forever to get there so aka I def need more practice.

    I solved for the first part V=V0+at
    =20m/s + (-10m/s2)(0.50s)
    =15m/s

    Then substituted that Vf value as the Vi value for the second part
    Vf2=V02 +2(a)(x)
    Vf2=15m/s2+2(-10m/s2)(35m)
    Vf2=225m/s + (-700m/s2)
    Vf=[itex]\sqrt{}475[/itex]
    Vf=21.79m/s which =22m/s

    yay, thanks for all your help!
     
  8. Sep 30, 2012 #7

    ehild

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    The driver does not brake for 0.5 s, and its initial speed is unknown, not 20 m/s. During that half second the speed is the unknown vmax. After 0.5 s, the distance from the deer is x=35-0.5vmax. You know that x=vmax2/(2a): vmax2/20=35-0.5vmax. That is a quadratic equation. Solve.


    ehild
     
  9. Sep 30, 2012 #8
    Okay now I'm just confused.
     
  10. Sep 30, 2012 #9
    deleting my post.
     
    Last edited: Sep 30, 2012
  11. Sep 30, 2012 #10

    ehild

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    Why?

    ehild
     
  12. Sep 30, 2012 #11

    ehild

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    Azizlwl: Read about the rules of this Forum here: https://www.physicsforums.com/showthread.php?t=414380



    ehild
     
  13. Sep 30, 2012 #12
    It's not a homework question btw, I'm studying for a quiz.

    But anyways, I'm confused because I'm unsure where you derived that second formula from. I do understand I am looking for the initial speed. What happened to the acceleration though, why aren't I using that? All my formulas have acceleration in them.

    I have, x=V0t+1/2at2 and V2=V02+ax
     
  14. Sep 30, 2012 #13

    ehild

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    Your formulas have acceleration if the object accelerates. But the problem says that the reaction time of the driver is 0.5 s, before stepping on the brakes. Without stepping on the brakes there is no deceleration, the car moves with constant velocity for 0.5 s after noticing the deer. Constant velocity means zero acceleration.

    ehild
     
  15. Sep 30, 2012 #14
    Okay, that was stupid on my part. Thanks for clarifying. I thought it was a two step question, but you're saying I can do it in one step with a quadratic equation?
     
  16. Sep 30, 2012 #15

    ehild

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    It is two steps: The first step is motion with uniform velocity for 0.5 s. If you denote the velocity by Vo, the car travels Vo*t distance and gets closer to the deer. The new distance is x=35-Vo*0.5. The second step is motion with uniform deceleration. At the end the velocity is zero: 0=Vo2+ax, and a=-10 m/s2.

    ehild
     
  17. Sep 30, 2012 #16
    Yeah, okay every time I work it out I get the wrong answer
     
  18. Sep 30, 2012 #17

    ehild

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    Sorry, it was my fault. Of course, 0=Vf2=V02+2ax, with a=-10 m/s2, as you wrote in your post #6.

    ehild
     
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