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