You are driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 10 m/s2 . how much distance is between you and the deer when you come to a stop?

A) Mind you before your reaction time, you had be going at a uniform speed 20m/s so for the reaction time of 0.5 seconds, you had covered a distance  of:

20m/s*0.5s = 10 m

For the second part which involved deceleration, using: 

v = u - at,  Noting that there is deceleration. 

u = 20m/s, v = final velocity = 0, a = -10m/s². 

Let us solve for the time.

v = u + at

0 = 20 - 10*t

10t = 20

t = 20/10 = 2 seconds.

Let us compute for the distance covered during the 2s

s = ut + 1/2at²,       a = -10 m/s²

s = 20*2 -0.5*10*2² = 20m

So the total distance covered = Distance covered from reaction time + Distance covered from deceleration
              
                                              = 10m + 20m = 30m

So you have covered 30m out of the initial 35m.

Distance between you and the dear:   35 - 30  = 5m

So you have 5m between you and the deer. So you did not hit the deer.

So the dear is 35m away from the car that is running at a speed of 20m/s where your reaction time in hitting the brakes is 0.5s and the cars deceleration is 10 m/s2. By calculating i came up with an answer of 5meters, the car only have 5 meters away from the deer to hit it.

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Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.