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You are driving down the highway and you brake to avoid hitting a deer

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data
    You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 m/s^2.

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


    2. Relevant equations

    I know we have to use both of these equations-

    x= x0 + v0t+ 1/2at^2
    and
    v=v0+at

    3. The attempt at a solution

    Reaction time eats up .5 of your seconds, so in the other half of a second, you are now 25 m from the deer.
    So for you, x=20t+1/2at^2

    And then your velocity-

    v= 20+at

    Then for the deer, since he's doing nothing, he would be

    x=25? Because he's not doing anything else but standing 25 feet away?

    And then his velocity wouldn't exist?

    So then what would you do to solve this?
     
  2. jcsd
  3. Aug 23, 2011 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, the deer is stationary. You don't really need to worry about its equations of motion.

    Solve for the value of x (the position of the car) at the time when v = 0. This tells you how far in front of the deer you are when you come to a stop.

    I would set x to be equal to your distance from the deer (who is at x = 0, the origin), so that x0 (the initial position, at t = 0) is +25 m. Don't forget that the formula is then:

    x = x0 + v0t + (1/2)at2

    If you take "forward" to be the negative direction (since your position is getting closer to 0 as you move forward), then your initial velocity is negative and your acceleration is positive (since it is rearward).

    You're free, of course, to choose the exact opposite sign convention (in which case your initial position is -25 m, your initial velocity is positive, and your acceleration is negative). Either way will work.

    I put my tip on how to actually solve the problem in italics. The rest is just helpful additional information.
     
  4. Aug 24, 2011 #3

    PeterO

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    Homework Helper

    I copy here a line from your solution:

    "Reaction time eats up .5 of your seconds, so in the other half of a second, you are now 25 m from the deer."

    Now the question

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


    Notice how the question was concerned about distances and speeds, while your first thought was about times??

    I think you should be using formulae which yield displacement and velocity rather than time.
     
  5. Aug 24, 2011 #4
    You're right, reaction time does mean constant velocity, which as you say, means that effectively deceleration doesn't start until 25m away. Then the problem starts:
    Distance to Deer: 25m
    Deceleration: -10 m s^-2
    Initial Velocity: 20 m s


    Using those equations, you're going to need to find how much time it will take to reach 0 velocity. So, equal the first equation to zero and you'll get a time value:
    v=u+at
    v=0=20+(-10t)

    Insert that value into the first equation and you'll get how far the car has travelled in t seconds. If you use the sign convention suggested by the user above, I would get slightly confused, but it's completely up to you. Just note that if you take the signs to be the ones I said, the distance travelled will be something positive, so you'll have to subtract it from 25 metres, because that's how far away the deer is. You may well find, that if the distance is more than 25, you get a negative answer. Not to worry, that just means the imaginary deer is dead, because you drove straight past it!

    (Actually, the equations would change once you hit the deer, since your deceleration will be larger, but not to worry about that! If you ignore the force exerted by hitting the deer, I get that you've gone 5 metres beyond the deer.)
     
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