You"re driving down the highway so late one night in ~ 20m/s as soon as a deer actions onto the road 42m in front of you. Your reaction time before stepping on the brakes is 0.50s , and also the preferably deceleration that your auto is 11m/s2 .

You are watching: What is the maximum speed you could have and still not hit the deer?

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

I understand you need to use the quadratic formula however i"m not sure what come plug in, have the right to someone please explain how to acquire the answer?

Thanks!

vf = v0 + at, the last 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 it is in displaced during the reaction time and also 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, friend should discover that:

v0 = 25.39 m/s i; this is the maximum rate that you space looking for.

A simple means to execute it would be to turning back the situation. If the auto is decelerating in ~ a preferably a= -11m/s^s, end a full distance of 42m, then you have the right to say, "what would certainly be the final velocity if the vehicle increases by 10m/s^2 over 42 meters?

You d use formula Vf^2 = 2(a)(deltaX) to discover V

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

Vf = 30.39 m/s.

Now you have actually to number in the .5s reaction time, so uncover the full time it require to accelerate to 30.39 m/s through an acceleration the 11 m/s^2.

Use formula a= deltaV/deltat

plug in what you know

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

t= 2.76s

now remember, girlfriend re decelerating because that 2.76s - .5s = 2.26s

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

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

DeltaV = 24.89 m/s. Now referring ago to the initial question, due to the fact that we know that the last speed is 0 m/s, then initial is 24.89 m/s.

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distance auto moves before brakes are used (reaction distance) = rd = (20)(0.50) = 10.0 m

distance automobile moves ~ 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 the takes to prevent from immediate deer is seen = 0.50 + 20/11 = 2.318 s

Vavg of you for the moment it takes to stop initially going at max speed = 28.2/2.318 ≈ 12.1 m/s

Your max rate at the prompt deer is seen in order come stop in time = 2(12.1) = 24.2 m/s ANS

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 - in ~ = v0

v0 = vf - at

v0 = 0 - 11(2.32)

v0 = 25.52 m/sMaximum allowable rate to no hit the deer

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