# Chasing Car Problem

1. Nov 9, 2008

### Axpyre

1. The problem statement, all variables and given/known data
A car travelling at 36 m/s passes a policeman who immediately accelerates at 3.01 m/s^2 from rest in pursuit. To the nearest second how long does it take the policeman to catch the speeder?

2. Relevant equations
I need help..

3. The attempt at a solution
I need major help..

2. Nov 9, 2008

### asleight

Set up kinematics for each of the drivers.

$$x_f=x_i+v_0t+0.50at^2$$.

3. Nov 9, 2008

### Axpyre

Can you explain that formula? I haven't learnt it yet. What is I?

4. Nov 9, 2008

### asleight

The final position of an object is equal to its initial position plus its initial velocity times the amount of time it travels plus one-half the acceleration times the time traveled squared.

You know how fast each of the cars were moving to begin with, that they begin at an initial position of zero, that they end up at the same place, and they have diff. accelerations--solve.

5. Dec 28, 2008

### NotaPhysicist

Yes that works. Work out kinematics for each then combine the equations and solve. I used it for similar problem finding the time it takes for a dog to catch up to a rabbit with 50m head start, given only the velocities of the dog and rabbit.

Very nice, thank you.

6. Dec 28, 2008

### FiskiranZeka

I'm also a student ( I think ) ;
after the moment thief and cop is next to each other;
if,
The Way The Thief Drives = The Way The Cop Drives
Then Cop catches the Thief.
36 . t = 0,5 . 3,01 . t^2
is the equation right ?

7. Dec 28, 2008

### asleight

Yes, that's exactly right.

8. Dec 28, 2008

### Gnosis

I used 3.0 m/s^2 rather than 3.01 m/s^2.

It would take 24 seconds for the police vehicle (starting from rest and accelerating uniformly at a rate of 3.0 m/s^2) to intersect with the pursued vehicle, which is traveling at a steady rate of 36 m/s. They would intersect 864 meters further down the highway.

½ x (3.0 m/s^2) x (24 seconds^2) = 864 meters (distance achieved by police vehicle in 24 seconds)

(36 m/s) x (24 seconds) = 864 meters (distance achieved by pursued vehicle in 24 seconds)

However, the police vehicle would have achieved far greater velocity by this point (in fact, 72 m/s), so it would simply pass the vehicle. If however, the police vehicle had to ram the vehicle Kamikaze style, then it would make contact in just 24 seconds.

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Here’s the time required per an acceleration of precisely 3.01 m/s^2:

½ x (3.01 m/s^2) x (23.92026578 seconds^2) = 861.1295681 meters (distance of accelerating police vehicle)

(36 m/s) x (23.92026578 seconds) = 861.1295681 meters (distance of speeding vehicle at steady velocity)

Since the time is the same for each equation, I used the ratio of the known values between both equations to establish the distance when the two vehicles would intersect. 36 m/s is known in the speeder’s equation while ½ x 3.01 m/s^2 (which equals 1.505) is known from the accelerating police vehicle equation therefore:

36 m/s / 1.505 = 23.92026578 seconds

23.92026578 seconds x 36 m/s = 861.1295681 meters (just as the kinematics of both equations agree)

Last edited: Dec 28, 2008