Question

tion when both trains have stopped? is (a 42 You are arguing over a cell phone while trailing an unmarked police car by 25 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 phoneW and yell, "I won't do that!"). At the beginning of that 2.0 s, the po- lice officer begins braking suddenly at 5.0 m/s2. (a) What is the sep- aration between the two cars when your attention finally returns? Suppose that you take another 0.40 s to realize your danger and begin braking. (b) If you too brake at 5.0 m/s, what is your speed when you hit the police car?

Answer 1

answer) a) we have vo=110km/h=31m/s

distance travelled in 2 s=31*2=62m

we have the formula

s-s_o=ut+1/2at²...............1)

s_o=0

s=31m/s*2+1/2*-5m/s²*2²=52m

so difference in distance=62-52=10m

and the separation between the two cars=25-10=15m=answer

b)total time=2+0.4=2.4s

distance travelled in this new time by me=31*2.4s=74.4m

distnace travelled by police car =31*2.4+1/2*-5*2.4²=60m

difference in distance=74.4-60=14.4m

and the distance between car 1and car 2=25-14.4=10.6m

now speed of the police car

v=u+at=31m/s+(-5m/s²)*2.4=19m/s

now using eqn 1 we have

for police car

s-10.6=19(t')+1/2*-5*(t')².............2)

for my car

s=31(t')+1/2*-5*(t')²............3)

substracting eqn 2 and 3 we have

10.6=(31m/s-19m/s)(t')

t'*12m/s=10.6m

t'=0.88s

we know the formula

v=u+at'

v=31m/s+(-5m/s)*0.88s=26.6m/s

v=26.6m/s or 27m/s.=answer.