# Homework Help: Have part 1 of problem stuck on part 2. help!

1. Sep 28, 2007

You are arguing over a cell phone while trailing an unmarked police car by 36 m. Both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.5 s (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that 2.5 s, the police officer begins emergency braking at 5 m/s2.

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

b) Suppose that you take another 0.4 s to realize your danger and begin braking. If you too brake at 5 m/s2, what is your speed when you hit the police car?

The following work is for part (a) which is correct. My work is shown below:
distance between car and police= 36m speed of cars v=110km/hr =110(5/18) = 30.55m/s
after police car breaks: initial v=20.55m/s, final v= 0m/s, a=-5m/s^2 from V^2-V^2-2as
0^2-30.55^2=2(-5)s s=933.64/10= 93.364 m

velocity of police car after 2.5 secs of breaking v=v+at 30.55+(-5)2.5=18.05 m/s
distance traveled in this time by police is S: (18.05)^2-(30.55)^2-2(-5)s =60.78m
distance traveled by car in 2.5s=velocity x time: 30.55 x 2.5= 76.375m
separation between 2 cars: (36+60.78)-76.375= 20.405m

For part (b) I did the following but its still incorrect. Any ideas?
v=u +at --> t=20.405/(30.55-18.05), t=1.6324s, from there I plugged the values into v=u+at 20.55-5(1.6324) and got 22.38775, rounded to 22.4 and then converted back to km/h and ended w/ 80.64 km/h which is still wrong.

2. Sep 28, 2007

### Staff: Mentor

Try this: Find the new separation of the two cars when you step on the brakes (after 2.9 seconds) and the speed of the police car at that moment. Then write the position equations for each, starting from that point, and solve for when they hit. Use that to find your speed at that point.

3. Sep 29, 2007

### Vijay Bhatnagar

This problem can also be solved using relative velocities and relative accelerations of the two vehicles. Let velocities and accelerations of "your" car and the police car be subscripted with 1 and 2 respectively. There are three distinct time periods - Period 1 = 2.5 sec (you are unmindful of the police car and only police car is under braking), Period 2 - 0.4 sec (you are mindful of the police car and only police car is under braking) and period 3 - when both vehicles are under braking. At time t=0 both cars are moving with same speed of 110 km/hr = 30.5 m/s.

At t=0, velocity of police car relative to your car = V21 = V2 - V1 = 0
At t=0, deceleration of police car relative to your car = a21 = a2 - a1 = -5 - 0 = -5 m/s^2.

Applying S = So + ut + 1/2 at^2 we get : Position of police car relative to your car = S21 = 36 + 0 - 1/2 x 5 x 2.5^2 = 20.4 m

During period 2 the motion continue as in period 1 (as the braking of your car is yet to begin). Distance between the two cars at the end of 2.9 sec = 36 + 0 - 1/2 x 5 x 2.9^2 = 15 m

Velocity of police car relative to your car at the end of this period (and beginning of period 3) = 0 - 5 x 2.9 = - 14.5 m/s

During period 3 both cars are under same deceleration. Hence, relative acceleration a21 = 0.
Displacement between the two cars at the beginning of this period = 15 m

Applying S = So + ut + 1/2 a t^2 we get : 0 = 15 - 14.5 t - 0 [ S = 0 when your car hits the police car] or t = 1.03 sec

At the beginning of this 1.03 sec period, velocity of your car = 30.5 m/s
Acceleration = - 5 m/s^2

Hence velocity of your car at the end of this 1.03 sec period = 30.5 – 5 x 1.03 = 25.35 m/s or 91.26 km/hr.