You're driving down the highway late one night at 16.0 m/s when a deer steps onto the road 39.0 m in front of you. your reaction time before stepping on the breaks is 0.50 s, and the maximum deceleration of your car is 12.0 m/s2 . how much distance is in between you and the deer when you come to a stop? what is the maximum speed you could have and still not hit the dear?

The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
The second part occurs after t = 0,5s with the given acceleration: a = - 12 m/s²
were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a

x₂ = 0.5*a*t² = 0.5*v°²/a

The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7

You can use this result to calculate the remaining distance. You can use the last equation to calculate the maximum speed you could have to avoid a collision.
Use x = 39m and solve for v₀. 

Electoral current force mechanical advantage if the wheel and axle is 2 n and the output force is 20 n