Question

You are arguing over a cell phone while trailing an unmarked police car by 40.0 m. Both your car and the police car are traveling at 108 km/h. Your argument diverts your attention from the police car for 2.50 s (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that 2.50 s, the police officer begins emergency braking at 5.00 m/s². ** (ANSWER PART (B) ONLY PLEASE)**

(a) What is the separation between the two cars when your attention finally returns?
24.375 m

(b) Suppose that you take another 0.400 s to realize your danger and begin braking. If you too brake at 5.00 m/s², what is your speed when you hit the police car?
__________________km/h

Answer 1

(A) in this time distance travelled by car = v t

= (108 x 1000 m / 3600 s) (2.50)

= 75 m

distance travelled by police car in this time,

= (108 x 1000 /3600)(2.50) - (5 x 2.50^2 / 2)

= 59.375 m

initial separation = 40 m

so final separation = 40 + 59.375 - 75

= 24.375 m .....Ans

(B) after 0.4 sec,

= 24.375 - (5 x 0.4^2 / 2)

= 23.975 m

suppose after time, car hits police car.

distance travel by you, d1 = (30t) - (5 t^2 / 2)

d1 = 30t - 2.5 t^2

at that time when you hit brakes, speed of police car

v =(30) - (5 x (2.50 +0.4) ) = 15.5 m/s

distance travelled by police car , d2 = 15.5t - 2.5 t^2

d1 - d2 = 23.975

(30 - 15.5)t = 23.975

t = 1.65 m/s

Speed of your car = 30 - (5 x 1.65)

= 21.73 m/s

in km/h = (21.73)(1/1000 km) / (1/3600 hr)

= 78.2 km/h .....Ans