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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.