Tricky Word Problem

1. Dec 14, 2008

kurtrudviking

1. The problem statement, all variables and given/known data

A motorcycle cop, parked on the side of a highway is passed by a woman in a red Ferrari 308 GTS doing 90km/h. After a few attempts to get his cycle started, the officer roars off 2.00s later. At what average rate must he accelerate if 110km/hr is the top speed and he is to catch her just at the state line 2.00km away?

2. Relevant equations

3. The attempt at a solution

I first started this question by showing what i know and making all unit conversions i need to.

1)
Cop for first 2 sec
vi= 0 km/hr
Pi=0 m
a=0
Ti= 0
Tf= 2

Cop after 2 sec
vi= 0 km/hr
vf= 110 km/hr
pf= 2000 m or 2 km
Ti= 2
a= X

Woman
vi= 90 km/hr
Pi= o m
a= 0
Tf=2

2) Convert the km/hr into m/s

(110 km/hr) X (1000 m/hr) X (1/60 Min) X (1/60 Sec) = 30.56 m/s (COP)
(90 km/hr) X (1000 m/hr) X (1/60 Min) X (1/60 Sec) = 25 m/s (woman)

3) Then find out how long it takes the Woman to get to the state line.

Woman
Position Final = Vi(T)
2000m = 25(T)
T = 80 sec

4) This means that the cop has 78 sec to catch her because it takes him 2 sec to start his bike.

thats all i got...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 14, 2008

Mthees08

We had a problem exactly like this in AP Physics my Sr. year. If I remember correctly your best bet is to find the womans condition after 2 seconds. Now use some algebra, and a constant acceleration equation (one with all variables x, v, a,) and set them equal, by either t or X it is your choice, however think about which will cancel out better. and you can use the border (or the limit as you approach it) as your final displacement.

Although the best answer is truly it doesnt matter, because it will take longer to solve the problem than to write the ticket. :)

3. Dec 14, 2008

kurtrudviking

thanks so much! much appreciated if anyone was curious this is how i ended up solving it.

Women Max = 25 m/s
Cop Max = 30.5 m/s

amount of time available
2000 m/25 m/s = 80 s - 2 s = 78 s

.5(a)(t1)^2 + 30.5(t2) = 2000 m
t1 + t2 = 78 s
a(t1) = 30.5

.5(30.5)(t1) + 30.5(78 - t1) = 2000
15.25(t1) + 30.5(78) - 30.5(t1) = 2000

t1 = (2000 - 30.5(78)) / (15.25 - 30.5) = 25.137 sec

a = 30.5 / 25.137 = 1.216 m/s^2

4. Dec 15, 2008

LowlyPion

This problem isn't all that tricky.

You have the first step already, which is determining the time budget. At 25 m/s the 2000 m can be covered in 80 seconds less the 2 second head start ... so 78 s.

Now look at the cop's velocity as a function of time.

There is an acceleration time call it T1 and time T2 at max speed.

You know his Vmax and you know for uniform acceleration his Vavg during acceleration will be half Vmax

To arrive at the Ferrari in 2 km then his distance traveled will need to be

2000 = 1/2*Vmax*T1 + Vmax*T2

won't it? And you already know T1 + T2 = 78

... so 2 equations and 2 unks.

5. Dec 15, 2008

kurtrudviking

thanks man both you guys helped me a ton :)
now that you guys showed me this it makes total sense and i cant believe i didnt think of it before. have a great day guys