Question

You are arguing over a cell phone while trailing an unmarked police car by 25.0 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 0.400 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?

Answer 1

(a)

lnitial distance b/w both cars = 25 m

initial speed of both the cars, vy = vp = 110 km/h = 30.55 m/s

where, vy = speed of your car

vp = speed of police car

For police car (Using kinematic equation),

Distance covered, xp = up*t - (1/2)at^2

xp = 30.55*2 - (1/2)*5.2*2^2

xp = 50.71 m

Distance covered by your car,

xy = 30.55*2 = 61.11 m

separation between the two cars,

d = (25 + 50.71) - 61.11

d = 14.6 m

(b)

you take another 0.400 s to realize your danger and begin braking,

t = 2.4 s

Distance covered by police car,

xp = 30.55*2.4 - (1/2)*5.2*(2.4)^2

xp = 58.34 m

distance covered by your car,

xy = 30.55*2.4 = 73.32 m

xp - xy = 14.98 m

distance b/w both cars,d = 25 - 14.98 = 10.02 m

Speed of your car at 2.4 s,

v = u - at

v = 30.55 - 5.2*2.4 = 18.07 m/s

Let, time required to hit the car = t

t = d / v

t = 10.02 / (30.55 - 18.07) = 0.802 s

your speed when you hit the police car,

v = 30.55 - 5.2*0.802

v = 26.37 m/s

v = 94.95 km/h