A simple way to do it would be to reverse the situation. If the vehicle is decelerating at a maximum a= -11m/s^s, over a total distance of 42m, then you can say, "what would be the final velocity if the vehicle accelerates by 10m/s^2 over 42 meters?

You d apply formula Vf^2 = 2(a)(deltaX) to find V

Vf = sqrt ((2)(11m/s^2)(42m)) = sqrt(924m^s/s^s)

Vf = 30.39 m/s.

Now you have to figure in the .5s reaction time, so find the total time it take to accelerate to 30.39 m/s with an acceleration of 11 m/s^2.

Use formula a= deltaV/deltat

plug in what you know

11 m/s^2 = (30.39m/s)/deltat, solve for t

t= 2.76s

now remember, you re decelerating for 2.76s - .5s = 2.26s

so find the velocity at 2.26s using equation a=(deltaV)/(deltat)

11m/s^2 = (deltaV)/2.26s.

DeltaV = 24.89 m/s. Now referring back to the original question, since we know that the final speed is 0 m/s, then initial is 24.89 m/s.

I think this is the correct answer for **What is the maximum speed you could have and still not hit the deer?**

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