OpenStudy (anonymous):

Situation question. You are arguing over a cell phone while trailing an unmarked police car by 25m; both your car and the police car are travelling at 110km/h. Your argument diverts your attention from the police car for 2.0s . At the beginning of that 2.0 ,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 5.0m/s^{2}, what is your speed when you hit the police car?

6 years ago
OpenStudy (anonymous):

This is a tricky one - but can be made simple by using concept of relative velocity. The initial condition is that both cars are travelling at 110kmph with a separation of 25m. In this condition, imagine that you are in the trailing car. The police car in front of you will appear stationary (due to zero relative velocity) at a distance of 25m. From this frame of reference - ie, the point of view of the careless driver, you can forget that both vehicles have a 110kmph speed already. Now suddenly, the police car brakes. From your POV, it will appear as if the stationary police car is now suddenly accelerating at you at 5 m/s^2. So, in 2 seconds, the separation reduces by: s = 1/2 * a * t^2 = 0.5 * 5 * 2 ^2 = 10m (reduction in separation) So, at the end of 2 seconds, the separation between you and the police car is 25 - 10 = 15m Applying the same logic, the separation at the end of 2.4s is: 25 - (1/2 * 5 * 2.4^2) = 10.6m But of course, the separation will continue to keep reducing unless the driver takes action.

6 years ago
OpenStudy (anonymous):

Now the tricky part: At the beginning of the 2 seconds (ie, when the police car starts braking), the police car appeared stationary to you. Now at the end of 2 seconds, the police car is closing in on you (from your POV). The velocity of that approach is at 2 seconds is: v = at = 5 * 2 = 10m/s. Now if you (the careless driver) don't take any action, the velocity of merge will keep on increasing. However, at 2 seconds, the driver starts braking at the same rate as the police car. Now, there is no relative acceleration (difference in acceleration) between the 2 cars. In such a case, the relative velocity of merging will continue the same. ie, the separation will continue to close in at a steady rate of 10m/s from that instant onwards. Reviewing the situation at the end of 2 secs (2s after the police car started braking and at the instant when you started braking): Separation = 15m (solved earlier) Velocity of merge = 10m/s (steady) So, the two cars will hit when the separation reduces to zero. That is in: 15m / 10 m/s = 1.5s. Now, you have been braking at a rate of 5m/s for 1.5s (from initial 110kmph) before you hit the police car. Thus your final velocity is: v = u - at = 110kmph - (5 * 1.5)m/s = 83 km/hr or 23.055 m/s

6 years ago