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?

+3
1. 24 February, 06:01
0
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.