- #1
Shahin_2010
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Homework Statement
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. 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?
I have a problem with part b) of this question, but I have provided the solution to part a) for completeness.
Homework Equations
Constant acceleration formulas.
The Attempt at a Solution
For part a) I set up the two cars relative to an axis, where your car is placed at the origin and the police care placed 25 m from the origin. Since you are traveling at constant speed (constant zero acceleration), I used x = v*t, to get x = 275/9* 2 = 550/9 m, (I converted the 110km/h to 275/9 m/s). So 'your car' travels a total of 550/9 m from the origin.
now to the police car, At 25 m from the origin the police is traveling at 275/9 m/s at a = -5 m/s^2, t = 2.0 s. Using v = u + at, where u = 275/9 m/s , t = 2.0 s, a = -5 m/s^2, We get v = 185/9 m/s. I then used v^2 = u^2 + 2as, to get s = 460/9 m. Remembering that the police car had traveled s = 460/9 m from 25 m away from the origin. Therefore the new distance ('s1')from the origin for the police car is s1 = s + 25 = 685/9 m.
Then I simply subtracted 685/9 - 550/9 = 15 m. The new separation between the two cars is 15m.
For part b) here is where I had some problems. Here is my reasoning, from the beginning of the 0.4 s to 0.4 s, 'your car' travels at 110 km/h (275/9 m/s) for t = 0.4 s, x = 275/9*0.4 = 110/9 m, therefore your car travels d = 550/9 + 110/9 = 220/3 m. This is about as far as I got, What should I calculate for, the new separation of the two cars, that traveled in the t = 0.4 s ?