Constant Acceleration: A Special Case

[Shahin_2010]

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 ?

 

[PhanthomJay]

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.

Yes, that looks right.

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 ?

Yes, you can do this in the same way you did part a, by using t = 2.4 instead of t=2.0. So at that instant after 2.4 seconds have elapsed, they are a certain distance apart. Now use apropriate kinematic equations for each vehicle, which collide each at the same distance away from where you started.

 

[Shahin_2010]

Thanks, a lot Jay.

Here is my solution, for 'your car' since I've got all the values; u = 275/9 m/s , a = - 5 m/s^2 , t = 2.4 s , s = ? using s = ut + 0.5at^2

we get s = 884/15 m.

since the police car traveled a d1 = 685/9 m in the first 2.0 s, and d2 = 185/9*0.4 = 74/9 m, in the next 0.4 s and a total of d = d1 + d2 = 253/3 m.

and then using v^2 = u^2 + 2as, I get v = 26 m/s (94 km/h).

One more thing Jay, could you give me your reasoning as to why you chose to use 2.4 s instead of 0.4 s for 'your car' ?

for eg: s = (275/9)(2.4) + 0.5(-5)(2.4)^2 what does the '0.5(-5)(2.4)^2' tell us ? is it an update that after 'your car' has traveled s = (275/9)(2.4), because this is where I get confused?

Kind Regards,
Shahin.

 

[PhanthomJay]

Thanks, a lot Jay.

Here is my solution, for 'your car' since I've got all the values; u = 275/9 m/s , a = - 5 m/s^2 , t = 2.4 s , s = ? using s = ut + 0.5at^2

we get s = 884/15 m.

No, you had it right the first time...the 'your car' is still moving at constant speed for the first 2.4 seconds, so the distance it has traveled in 2.4 seconds is vt = (275/9)(2.4) = 73.3 m (that's the 220/3 m answer you came up with originally).

since the police car traveled a d1 = 685/9 m in the first 2.0 s, and d2 = 185/9*0.4 = 74/9 m, in the next 0.4 s and a total of d = d1 + d2 = 253/3 m.

No, the police car has been decelerating since t = 0 when you started to not pay attention. It has thus traveled s = s_o + v_o(t) -1/2 (5)(2.4)^2 = 25 + (275/9)(2.4) - 14.4 = 83.9 m from wher your car was at the the start time t =0. So at this point in time 2.4 seconds later, they are 83.9 - 73.3 = 10.6 m apart.Your car is still traveling at 275/9 m/s (30.65 m/s) and the police car is traveling at v = 275/9 -5(2.4) = 18.5 m/s. Now their accelerations are the same from this point forward up to the crash. So now let's start again from this point. Your car has an initial speed of 30.65, an acceleration of -5 m/s/s, and it is 10.6 m behind the police car. The police car at this pooint has an initial speed of 18.5 m/s, an acceleartion of -5 m/s/s. When they crash, it will be at the same elapsed time and same distance from the start. Continue...