**You're driving down the highway late one night at 20 m/s when a deer steps onto the road 49 m in front of you. You reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10m/s2.**A. How much distance is between you and the deer when you come to a stop?

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

**Answer:**

**v = 26.7 m/s**

**Explanation:**

Given,

speed of the car = 20 m/s

distance between the car and the deer = 49 m

time taken to press the brake = 0.50 s

maximum deceleration of the car = 10 m/s²

Now,

distance travel by the car in 0.5 s = u x t = 20 x 0.5 = 10 m

distance travel by the car after the break is pressed

Using equation of motion

v² = u² + 2 a s

0² = 20² - 2 x 10 x s

s = 20 m

Total distance travel by the car = 20 + 10 = 30 m

Distance between deer and car = 49-30 = 19 m.

b. Maximum speed a car could have

Distance travel by the car in reaction time = v' x 0.5

v' is the maximum speed of the car.

maximum distance car can cover = 49 - 0.5 v'

Now, Using equation of motion

v² = u² + 2 a s

0² =v'² - 2 x 10 x (49- 0.5 x v')

v'² +10 v' -980 = 0

By solving

**v = 26.7 m/s**

Hence, maximum speed of the car can be 26.7 m/s

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Find another answersPhysics, published 10.03.2023