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. 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?

a) Between you and the deer, there is 5 m when you stop.

b) You could travel at 22 m/s and still not hit the deer.

Explanation:

Hi there!

The position and velocity of the car can be calculated using the following equations:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

When the velocity is constant, the equation of the position is:

x = x0 + v · t

a)You travel for 0.50 s before you apply the brakes. The traveled distance in that time can be calculated using the equation of position (let´s place the origin of the frame of reference at the point you see the deer):

x = x0 + v · t

x = 0 m + 20 m/s · 0.50 s

x = 10 m

You travel 10 m before you start to decelerate. Now let´s find the time it takes you to stop (to reach a velocity of 0 m/s). Using the equation of velocity:

v = v0 + a · t

0 = 20 m/s - 10 m/s² · t  (Notice that the acceleration is in the opposite direction of movement and, therefore, is negative). Let´s solve the equation for "t".

-20 m/s = -10 m/s² · t

t = -20 m/s / -10 m/s²

t = 2.0 s

Now we can calculate the distance traveled in 2.0 s. Notice that the initial position will be the distance traveled in the 0.5 s before you apply the brakes:

x = x0 + v0 · t + 1/2 · a · t²

x = 10 m + 20 m/s · 2.0 s - 1/2 · 10 m/s² · (2.0 s)²

x = 30 m

Between you and the deer, there is (35 m - 30 m) 5 m when you stop.

b) We have the following system of equations:

x0 = v0 · 0.50 s    (the distance traveled before breaking)

0 = v0 - 10 m/s² · t   ( the velocity equation when you stop)

35 m = x0 + v0 · t - 1/2 · 10 m/s² · t²   (the distance traveled until you stop)

Using the second equation:

v0 = 10 m/s² · t

Replacing v0 in the first equation:

x0 = 10 m/s² · t · 0.50 s

Replacing v0 and x0 in the third equation:

35 m = 10 m/s² · t · 0.50 s + 10 m/s² · t · t - 1/2 · 10 m/s² · t²

35 m = 5 m/s · t + 10 m/s² · t² - 5 m/s² · t²

35 m = 5 m/s · t + 5 m/s² · t²

0 = 5 m/s² · t² + 5 m/s · t - 35 m

Solving the quadratic equation:

t =2. 2 s

The maximum speed will be:

v0 = 10 m/s² · t

v0 = 10 m/s² · 2.2 s

v0 = 22 m/s

You could travel at 22 m/s and still not hit the deer.