Trailing Car: Separation & Speed Calculations

[josh84]

Homework Statement

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

Homework Equations

Sorry! I have no idea which equations do I have to apply to...

The Attempt at a Solution

Sorry! please help me to find solution...

 

[learningphysics]

In these type of problems, you want an equation that gives you position (ie displacement) in terms of time... so look for those types of kinematics equations...

Also, you need to decide when your t=0 time is... when is it convenient to set t=0... for this case I think the best time is right when the two cars are separated by 25m and the police car begins braking...

So at t = 0, your car is at x =0 and the police car is at x = 25... try to find an equation for the position of your car in terms of time...

Try to find an equation for the police car also...

 

[m2003]

As a scientist, it is important to approach problems systematically and logically. In this scenario, we are given information about the separation between two cars and their speeds, as well as the acceleration of one of the cars. To solve this problem, we can use the equations of motion, specifically the equations for constant acceleration.

(a) To find the separation between the two cars when the person's attention returns, we can use the equation:

s = ut + 1/2 at^2

Where s is the separation, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the initial velocity (u) is 110km/h, which can be converted to meters per second (m/s) by multiplying by 1000/3600. This gives us an initial velocity of 30.56 m/s. The acceleration (a) is given as 5.0 m/s^2 and the time (t) is 2.0 seconds.

Substituting these values into the equation, we get:

s = (30.56 m/s)(2.0 s) + 1/2 (5.0 m/s^2)(2.0 s)^2

s = 61.12 m + 10 m

s = 71.12 m

Therefore, the separation between the two cars when the person's attention returns is 71.12 meters.

(b) To find the person's speed when they hit the police car, we can use the equation:

v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

In this case, the initial velocity (u) is 30.56 m/s (as calculated above), the acceleration (a) is 5.0 m/s^2, and the distance (s) is the separation calculated in part (a), which is 71.12 meters.

Substituting these values into the equation, we get:

v^2 = (30.56 m/s)^2 + 2(5.0 m/s^2)(71.12 m)

v^2 = 935.94 m^2/s^2 + 710.6 m^2/s^2

v^2 = 1646.54 m^2/s^2

Taking the square root of both sides, we