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?

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.