You are arguing over a cell phone while trailing an unmarked police car by 26.0 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 5.20 m/s2.

(a) what is the separation between the two cars when your attention finally returns? suppose that you take another 0.500 s to realize your danger and begin braking.
(b) if you too brake at 5.20 m/s2, what is your speed when you hit the police car?

a) d = 15.6m

b) v= 26.44m/s

Step-by-step explanation:

Distance = 26.0m

Traveling rate = 110km/hr

=( 110*1000)/ 3600

= 30.5 m/s

Deceleration (-a) = 5.20 m/s^2

Let xi =initial position

xf = final position

Vi= initial velocity

Vf = final Velocity

a) Assume that my car starts at position xi = 0

The police car starts at circle = 26m

The initial distance between both cars = 26 - 0 = 26.0m

The position of my car after I gained my focus is given by

xf = xi + Vi(t)

= 0 + (30.5)2

= 61m

The position of the police car after I gained my focus is given as

xf = xi + Vi(t) + 1/2at^2

= 26 + (30.5)2 + 1/2(-5.2)(2^2)

= 26 + 61 - 10.4

= 87 - 10.4

= 76.6m

The separation distance after I regard my focus is 76.6 - 61

= 15.6m

D = 15.6m

b) Suppose it took another 0.50 secs to realize danger and begin to press brake, the initial position of my car (xi) = 61m

The initial position of the police car(xi) = 76.6m

The distance I traveled is given by

xf= xi + Vi(t)

= 61 + (30.5)0.5

= 61 + 15.25

= 76.25

Velocity the police car reached while decelerating when I wasn't focusing is given as

Vf= Vi + at

= 30.5 + (-5.2)2

= 30.5 - 10.4

= 20.1 m/s

This velocity is now the initial velocity of the police car. The position of the police car after 2 secs is given as

xf = xi + Vi(t) + 1/2at^2

= 76.6 + (20.1)0.5 + 1/2(-5.2)(0.5^2)

= 76.6 + 10.05 - 0.65

= 87.1 - 0.65

= 86.45m

The distance between the two cars became 86.45 - 76.25

= 10.2m

Velocity of the police car became

Vf = Vi +at

= 20.1 + (-5.2)0.5

= 20.1 - 2.6

= 17.5m/s

To know the time at which the two cars collide, set the position of the two cars equal to each other

xf(1) = xf(2)

xi(1) + Vo(t) + 1/2at^2 = xi(2) + Vi(t) + 1/2at^2

xi(1) + Vo(t) = xi(2) + Vi(t)

76.25 + (30.5)t = 86.45 + (17.5)t

30.5t - 17.5t = 86.45 - 76.25

13t = 10.2

t = 10.2/13

t = 0.78secs

Velocity of my car is given by

V= Vi +at

= 30.5 + (-5.2)(0.78)

= 30.5 - 4.056

= 26.44 m/s

answer: y=0 if that is what u r looking for

answer: the proof is mentioned below.

step-by-step explanation:

since, given:   lines bc = ef

prove:   lines ac > ef

          statements                                           reason

1. bc=ef                                           1. given

2. ac= ab+bc                                 2. because b is the point between the  

                                                            line segment ac

3.   ac > bc                                       3. because if a line segment is equal to

                                                          the sum of two line segments then it is

                                                          always greater then each of these

                                                          individual line segments

4. ac > ef                                         4. by equality property.