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1D Kinematics problem

  1. Aug 30, 2005 #1
    A motorist is travelling at 10 meters a second when he sees a deer in the road 50m ahead. If the maximum negative acceleration of his vehicle is -7 meters per second per second, then what is the maximum reaction time delta t of the motorist that will allow him to avoid hitting the deer?

    I am using the equation:

    change in x = initial velocity * time + (1/2) * acceleration * time squared

    or:

    x=v0*t+(1/2)*a*t^2

    I plug all of the information in (v0 = 10, x = 50, a = -7):

    50=10t-3.5t^2

    ... and I find no real solution. Am I doing this problem correctly and if so what does this mean in terms of a numerical answer for the reaction time of the motorist? I'm a physics dumbhead so please have mercy upon me and thanks in advance :)
     
  2. jcsd
  3. Aug 30, 2005 #2

    Doc Al

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    Staff: Mentor

    The problem is not so much the physics as it is decoding these kinds of word problems. By "reaction time" is meant: How long can the driver delay hitting the brakes and still stop before hitting the deer?

    Start by finding the stopping distance. (Remember that equation you derived in your earlier question? It's going to come in handy here.)
     
  4. Aug 30, 2005 #3
    Can I ask what you mean by the stopping distance?
     
  5. Aug 30, 2005 #4

    Doc Al

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    Sure. "stopping distance" = From the moment you step on the brakes, how far do you travel before coming to a stop.

    You are given the maximum acceleration and the initial speed. Dig up that formula that relates speed, distance, and acceleration. Use it to figure out the stopping distance.

    Since the deer is 50m away, let's hope the stopping distance turns out to be something less than 50m! (For instance, if it takes 30m to come to a stop, then you know that the driver can coast for another 20m before hitting the brakes.)
     
  6. Aug 30, 2005 #5
    I can't find anything better in the book than the equation:

    vf^2=v0^2+2*a*(x-x0)

    This seems worthless to me since I don't know what the final velocity is. :/ I have no idea how to find the stopping distance (which I assume is x-x0 in this problem rather than just x0).
     
  7. Aug 30, 2005 #6

    Doc Al

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    That's the one you want.

    Here's a hint: It comes to rest!
    Don't be so quick to throw in the towel. You have to wrestle with these problems, trying and making mistakes until you get it. There is no other way. (A problem that you know immediately how to solve is too easy!)
     
  8. Aug 30, 2005 #7
    Ok if it comes to rest, then the final velocity would be 0. This would give us the equation:

    0=10^2+2*-7*(50-x0)

    When I sovle this for x0, I get -42.85 roughly. This would mean the stopping distance is 50--42.85 or about 92.8 meters, but how can that be greater than 50 meters (distance to deer) and of what use is this number?
     
  9. Aug 30, 2005 #8

    Doc Al

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    Careful. The "stopping distance" is what you are calling "50 - x0". Let's go back to your original equation:
    vf^2=v0^2+2*a*(x-x0)

    For simplicity, let's call the initial position to be zero (x0 = 0). Thus, since vf = 0, we get:
    0 = v0^2 + 2*a*x

    Plug in your numbers and solve for x; that will give you the stopping distance.
     
  10. Aug 30, 2005 #9
    Thanks! :approve:

    I managed to solve the problem by getting the stopping distance and subtracting it from the total distance (given in problem) to find the first part of the distance (before he breaked) and using the given velocity to solve for his "reaction time".

    Do you have any advice on solving physics problems in general and figuring out what the problem is asking for?
     
  11. Aug 30, 2005 #10

    Doc Al

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    Excellent!

    This guy has some good general advice: http://www.oberlin.edu/physics/dstyer/SolvingProblems.html

    But the only way to learn physics problem solving is to solve as many problems as possible, and don't settle for just "getting the answer" but think of alternate ways of solving the same problem. Often there are multiple ways of looking at the same problem.

    And don't be surprised if some problems take hours (sometimes days) to crack because you're just not looking at it right or whatever. There are no shortcuts! But the more problems you solve, oddly enough, the "smarter" you will get. :wink:

    That and "Don't Panic!". :smile:
     
  12. Aug 30, 2005 #11
    Good book, on an off topic note. I wonder what it has to say about math-challenged undergraduates :?
     
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