You’re 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. 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?

(a) Distance between deer and car = 5 m

(b) Vmax = 21.92 m/s

Explanation:

a.

First we calculate distance covered during response time:

s₁ = vt   equation 1

where,

s₁ = distance covered during response time = ?

v = speed of car = 20 m/s

t = response time = 0.5 s

Therefore,

s₁ = (20 m/s)(0.5 s)

s₁ = 10 m

Now, we calculate the distance covered by the car during deceleration. Using 3rd equation of motion:

2as₂ = Vf² - Vi²

s₂ = (Vf² - Vi²)/2a eqation 2

where,

a = deceleration = - 10 m/s²

s₂ = Distance covered during deceleration = ?

Vf = Final Velocity = 0 m/s (since car finally stops)

Vi = Initial Velocity = 20 m/s

Therefore,

s₂ = [(0 m/s)² - (20 m/s)²]/2(-10 m/s²)

s₂ = (400 m²/s²)/(20 m/s²)

s₂ = 20 m

thus, the total distance covered by the car before coming to rest is given as:

s = s₁ + s₂

s = 10 m + 20 m

s = 30 m

Now, the distance between deer and car, when it comes to rest, can be calculated as:

Distance between deer and car = 35 m - s = 35 m - 30 m

Distance between deer and car = 5 m

b.

Since, the distance covered by the car in total must be equal to 35 m at maximum. Therefore,

s₁ + s₂ = 35 m

using equation 1 and equation 2 from previous part:

Vi t + (Vf² - Vi²)/2a = 35 m

Vi(0.5 s) + [(0 m/s)² - Vi²]/2(-10 m/s²) = 35 m

0.5 Vi + 0.05 Vi² = 35

0.05 Vi² + 0.5 Vi - 35 = 0

solving this quadratic equation, we get:

Vi = - 31.92 m/s  (OR)  Vi = 21.92 m/s

For maximum velocity:

Vmax = 21.92 m/s

the results is a decrease in air temperature with an increase of altitude.

explanation: