You are arguing over a cell phone while trailing 4


You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0 s (long enough for you to look at the phone and yell, “I won’t do that!”). At the beginning of that 2.0 s, the police officer begins braking suddenly at 5.0 m/s^2. (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another 0.40 s to realize your danger and begin braking. (b) If you too brake at 5.0 m/s^2, what is your speed when you hit the police car?

Solution:

Let us first convert 110 km/h into m/s, which gives us, 30.56 m/s.

 
a) In 2 seconds, you travel, 30.56m/s\times 2s=61.11m. To find how far the police car travels during this time, we use the following equation:

x-x_0=v_0t+\dfrac{1}{2}at^2

(where v_0 is initial velocity, x is final displacement, x_0 is initial displacement, t is time and a is acceleration.)

Substituting the values given in the question gives us:

x=(30.56m/s)(2.0s)+\dfrac{1}{2}(-5.0m/s^2)(2.0s)^2

x=51.11m

This means the gap between the two cars has narrowed by 10.0 m (61.1 m – 51.11 m=10.0 m) which means the distance between the cars is now 15.0 m (25 m – 10 m=15 m).

 

b) As the question states that it takes us an additional 0.4 seconds to realize our danger and begin braking, we will add this to the total time, giving us a total time of 2.4s. During this 2.4 seconds, we travel 73.33 m and the police car travels 58.93 m (using the same equation as above, and same values except for a new time value of 2.4s). Now, the initial distance between the cars has narrowed by 14.4 m. This means at the start, when you started to break, the gap between the cars is 10.6 m. To calculate the speed of the police car at this time, we can use the following equation:

v=v_0+at

v=30.56-5(2.4)

v= 18.56m/s

 

It must be noted that the collision between the cars will occur when x_{you}=x_{police} and we will choose x_{you}=0 to be the reference position. This means the position of the police car is, x_{police}=10.6m.

 

Now, we can write two equations, (same equation used in part a):

x_{police}-10.6m=18.56m/s(t-t_0)-\dfrac{1}{2}(5m/s^2)(t-t_0)^2

x_{you}=30.56m/s(t-t_0)-\dfrac{1}{2}(5m/s^2)(t-t_0)^2.

 

We can subtract the equations from each other, which would lead us to:

10.6m=(30.56m/s-18.56m/s)(t-t_0)

0.883s = t-t_0

 

At this time, your speed is:

30.56m/s+a(t-t_0)=30.56m/s-5m/s^2(0.883s)=26m/s which is approximately, 94 km/h.

 

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 42.

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