Question

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 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 11 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?

Answer 1

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

a)

• In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
• The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
• We can find this distance simply applying the definition of average velocity, as follows:

       $\Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)$

• The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
• We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

       $v_{1f} ^{2} - v_{1o} ^{2} = 2* a* \Delta x (2)$

• where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
• Solving for Δx, we get:

       $\Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)$

• So, the total distance traveled was the sum of (1) and (3):
• Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
• Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

• We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
• The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
• ⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
• Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

       $\Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)$

• Δx₂, is the distance traveled while decelerating, and can be obtained  using (2):

        $v_{omax} ^{2} = 2* a* \Delta x_{2} (7)$

• Solving for Δx₂, we get:

       $\Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a} = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)$

• Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
• Taking the positive root in the quadratic formula, we get the following value for vomax:
• v₀max = 24.3 m/s.