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What is the maximum speed you could have and still not hit the deer?

You're driving down the highway late one night at 20m/s when a deer steps onto the road 42m in front of you. Your reaction time before stepping on the brakes is 0.50s , and the maximum deceleration of your car is 11m/s2 .

What is the maximum speed you could have and still not hit the deer?

I know you have to use the quadratic formula but i'm not sure what to plug in, can someone please explain how to get the answer?

Thanks!

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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.

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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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distance car moves before brakes are applied (reaction distance) =  rd = (20)(0.50) = 10.0 m

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
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vf = v0 + at

vf - v0 = at

(vf - v0)/a = t

t = (vf - v0)/a

t = (0 - 20)/-11

t = -20/-11

t = 1.82 s

Adding in the driver's reaction time:

t = 0.50 s + 1.82 s = 2.32 s

vf - at = v0

v0 = vf - at

v0 = 0 - 11(2.32)

v0 = 25.52 m/s Maximum allowable speed to NOT hit the deer

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