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24 February, 05:50

You're driving down the highway late one night at 16m/s when a deer steps onto the road 44m 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.

How much distance is between you and the deer? What is the maximum speed you could have and still not hit the deer?

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  1. 24 February, 06:01
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    Have you considered hitting it one piece at a time?

    Reaction Distance: Speed * 0.50 s

    Initial Problem: 16 m/s * 0.50 s = 8 m

    Braking Acceleration: a (t) = - 10 m/s^2 - -

    This is constant.

    Braking Velocity: s (t) = - 10 m/s^2 (t) + "Initial Velocity"

    Initial Problem: s (t) = - 10 m/s^2 (t) + 16 m/s

    Braking Location: x (t) = - 5 m/s^2 (t^2) + 16 m/s (t) - "Initial Location"

    Initial Problem: x (t) = - 5 m/s^2 (t^2) + 16 m/s (t) - (44 m - 8 m)

    Having said all that, the first question makes no sense. Does it mean the distance when the vehicle stops?

    s (t) = - 10 m/s^2 (t) + 16 m/s = 0 = = > t = 16/10 s = 8/5 s

    That's how long it takes to stop.

    Where are we when that happens? x (8/5) = - 5 m/s^2 ((8/5 s) ^2) + 16 m/s (8/5 s) - (44 m - 8 m) = - 23.2 m

    This seems to be a little different from your response. Now, the challenge is to do this all over again, but missing the initial velocity.

    Initial Velocity: V - - Let's just call it this so we can talk about it.

    Initial Distance is the same: 44 m - - No change.

    Reaction Distance: V/2

    Braking Acceleration: a (t) = - 10 m/s^2 - -

    No Change Braking Velocity: s (t) = - 10 m/s^2 (t) + V - - Missing V!

    Braking Location: x (t) = - 5 m/s^2 (t^2) + V (t) - (44 m - V/2 m)

    When do we stop? s (t) = - 10 m/s^2 (t) + V = 0 = = > t = V/10 s

    Where do we stop? x (t) = - 5 m/s^2 ((V/10 s) ^2) + V (V/10 s) - (44 m - V/2 m) = 0 - - We might just brush the poor, frightened deer.

    One piece at a time! Slowly. Methodically.
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