# Two cars, one accelerating and one

1. Sep 15, 2004

### Omid

The driver of a pink Cadillac traveling at a constant 60 mi/h is being chased by the law. The police car is 20 m behind the perpetrator when it too reaches 60 mi/h, and at that moment the officer floors the gas pedal. If her car roars up to the rear of the Caddilac 2.0 s later, what was her acceleration, assuming it to be constant.

Can you help me with this problem?
I have the answer and the solution, after reading that I thought understood it. But when started to solve the next problem, exactly the same type, I couldn't do anything.
Please help me understand this type of problems.
Thanks

2. Sep 15, 2004

### Tide

Look at it from the perspective of the Cadillac. The police car is accelerating at a constant rate (starting from rest) to cover a fixed distance. Use the formula distance = acceleration X time^2 /2

3. Sep 15, 2004

### nolachrymose

I thought it was D = V_i*t + A*t^2/2, since the initial velocity was 60m/h?

4. Sep 16, 2004

### HallsofIvy

Staff Emeritus
Yes, Tide, you missed it! The problem says that initially the police car was 20 feet behind the cadillac and also doing 60 mph when it starts accelerating.

5. Sep 16, 2004

### Omid

But I already had the answer, what I need is some explanations to understand the problem; NOT just a formula.
I don't really get it. Why do we use their relative speed? Why do we look at it from the perspective of the Cadillac.
I have 5 problems all the same type, A moving at a constant speed and B accelerating at a constant rate; find acceleration or time or displacement … Thus I'm sure one of them will be in our final exam. So I need to understand them very well.
Please help me, PLEASE
Thanks

6. Sep 16, 2004

### Staff: Mentor

Tide didn't miss anything. He is looking at things from a frame moving with the cadillac. It's actually the easy way to solve this. Since with respect to the cadillac the initial speed of the police car is 0 when it's 20 m behind.

7. Sep 16, 2004

### Staff: Mentor

So how did you solve it?
You don't have to do it that way. Another way is to write expressions for the position of the cadillac and the police car as a function of time. The cadillac has a constant speed, while the police car accelerates. You know they must be at the same position in 2.0 seconds... so set it up and solve for the acceleration.

8. Sep 16, 2004

### HallsofIvy

Staff Emeritus
Yes, you don't need to use "relative" speed. I didn't when I looked at the problem and didn't realize that Tide was working "relative to the cadillac".

The more dificult part of the problem is converting from "miles per hour" to "meters per second". Let the cadilac's speed be V m/s. Taking the 0 point at the cadilac's initial position, its position at time t second is x= Vt. The police car's position after t seconds is y= Vt+ (1/2)at2- 20. The police care catches up to the cadillac when Vt= Vt+ (1/2)at2- 20. Since you are given that the time to catching up to the cadillac is 2 seconds, let t= 2 and solve for a.

Notice that the "VT" terms in Vt= Vt+ (1/2)at2- 20 cancel out! THAT'S why you get exactly the same answer taking V to be 0: that would be "in a frame moving with the cadillac" with the police cars speed "relative to" the cadillac.

9. Sep 16, 2004

### Omid

I didn't, the answer was in teh solutions manual

10. Sep 18, 2004

### Omid

I don't know, what made me think that problem is a "new and unknown " problem. Maybe some mistakes in calculations and some misleading hints in the solutions manual.
Ok, thank you very much for clearing things up.

For now let me bring another " new and unknown " problem

A motorcycle cop, parked at the side of a highway, is passed by a red Ferrari doing 90.0 km/h. The officer roars up after 2.0 seconds. At what average rate must he accelerate to his top speed 110 km/h to catch her
2.0 km away? Consider that his top speed is 110 km/h.

Again two car, A accelerating and B with a constant rate.
The most important difference between this and the previous, I think , is the top speed given, 110 km/h.
It means we have two time intervals. The first in which the cop must accelerate and the second in which his speed is constant.
So there will be 3 unknowns. The cop's acceleration, the first time interval and the second.
If so, what are the formulas relating that quantities?
Thanks

11. Sep 19, 2004

### Staff: Mentor

So you know the distance the cop must travel (2.0 km). And you can find the time he has to get there (how long does it take the Ferrari to travel that distance?).

Yes, think of it as two time intervals, $t_1$ and $t_2$. Now find expressions, using these times, for the total distance traveled and the total time. Once you figure out the times, you can figure the acceleration.

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