The Best Answer for What is the maximum speed you could have and still not hit the deer?
The time it takes for the car to slow down from its maximum speed is given by:
vf = v0 + at, the final velocity vf is 0 m/s, the maximum initial speed is v0, and the acceleration a is -11 m/s^2.
0 = v0 – 11t => t = v0 /11
The car will be displaced during the reaction time and the slowdown time; therefore,
xf = x0 + v0(0.50 + v0/11) + 0.5at^2, the final position xf is 42 m and the initial position x0 is 0 m.
42 = v0(0.50 + v0/11) -5.5(v0/11)^2
42 = v0/2 + (v0^2)/11 – (v0^2)/22
42 = v0/2 + (v0^2)/22
If you use the quadratic formula correctly, you should find that:
v0 = 25.39 m/s i; this is the maximum speed that you are looking for.
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
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?
distance car moves after brakes are applied = (braking distance) = bd =1/2at² = (0.5)(11)(20/11)² = 18.2 m
total distance traveled by you from instant deer is seen = 28.2 m
total time it takes to stop from instant deer is seen = 0.50 + 20/11 = 2.318 s
Vavg of you for the time it takes to stop initially going at max speed = 28.2/2.318 ≈ 12.1 m/s
Your max speed at the instant deer is seen in order to stop in time = 2(12.1) = 24.2 m/s ANS