Physics 2.17?

Question

You're driving down the highway late one night at 18 m/s when a deer steps onto the road 40m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10m/s^2

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

How much distance is between you and the deer when you come to a stop?

Answer 1

Your working formula is

Vf^2 - Vo^2 = 2as

where

Vf = final velocity of car = 0 (when it stops)

Vo = velocity of car so it will not  hit the deer

a = acceleration = 10 m/sec^2

s = stopping distance so car will not hit the deer

NOTE -- since there was a 0.50 sec reaction time before brakes were applied, the car then moved a distance of 18 * 0.5 = 9 meters

Therefore, effectively, the stopping distance between the car and the deer is 40 - 9 = 31 meters.

Substituting appropriate values in the above formula,

0 - Vo^2 = 2(-10)(31)

NOTE, as well, the negative sign attached to the acceleration. This simply implies that the car was slowing down when the brakes were applied.

Vo^2 = 620

Vo = 24.90 m/sec.  --- maximum speed of car and still not hit the deer

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With the car's given speed of 18 m/sec, it will stop according to the formula

Vf^2 - Vo^2 = 2as

where all the terms have been defined above.

Substituting values,

0 - 18^2 = 2(-10)(s)

s = 16.2 meters

Since the car was 31 meters in front of the deer when the brakes were applied and stopped after travelling another 16.2 meters, then the distance between the car and the deer when the car stopped ---

31 - 16.2 = 14.8 meters

Answer 2

1.

vf^2 = vi^2 + 2ax

0 = vi^2 + 2(-10)(31)

2.

vf^2 = vi^2 + 2ax

0 = 18^2 + 2(-10)x

x=16.2

40 - 16.2 - 9 = 14.8m