Two cars, one accelerating and one

In summary, the conversation discusses a problem involving a police car chasing a pink Cadillac. The police car is initially 20 meters behind the Cadillac, both traveling at a constant speed of 60 miles per hour. The officer then accelerates and catches up to the Cadillac in 2 seconds. The conversation discusses different ways to solve the problem, including looking at it from the perspective of the Cadillac and using formulas to calculate acceleration. Another problem is also discussed involving a motorcycle cop chasing a red Ferrari, with a top speed of 110 km/h, and the need to find the cop's acceleration to catch up to the Ferrari 2.0 km away. The conversation also mentions the use
  • #1
Omid
182
0
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
 
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  • #2
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
I thought it was D = V_i*t + A*t^2/2, since the initial velocity was 60m/h?
 
  • #4
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
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 :cry:
Thanks
 
  • #6
HallsofIvy said:
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.
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
Omid said:
But I already had the answer, what I need is some explanations to understand the problem; NOT just a formula.
So how did you solve it?
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.
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
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
Doc Al said:
So how did you solve it?

I didn't, the answer was in teh solutions manual :biggrin:
 
  • #10
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
Omid said:
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.
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?).


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?
Yes, think of it as two time intervals, [itex]t_1[/itex] and [itex]t_2[/itex]. 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.
 

What is the difference between acceleration and speed?

Acceleration is the rate of change of an object's velocity over time, while speed is the measure of how fast an object is moving at a particular moment in time. In simpler terms, acceleration is how quickly an object's speed is changing, while speed is how fast the object is currently moving.

How do you calculate acceleration?

Acceleration can be calculated by dividing the change in an object's velocity by the time it takes for that change to occur. The formula for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. This formula can be applied to calculate acceleration for both cars in this scenario.

Can an object have constant speed and still be accelerating?

Yes, an object can have constant speed and still be accelerating. This is because acceleration takes into account the change in an object's velocity, not just its speed. For example, if a car is driving at a constant speed of 60 miles per hour but then suddenly changes direction, it is still accelerating even though its speed remains constant.

What factors affect the acceleration of a car?

The acceleration of a car can be affected by several factors, including the force applied to the car (such as pressing on the gas pedal), the mass of the car, and any external forces acting on the car (such as friction or air resistance). The type of surface the car is driving on can also affect its acceleration.

How does acceleration impact fuel efficiency?

The acceleration of a car can impact its fuel efficiency. The faster a car accelerates, the more fuel it will use. This is because accelerating requires more energy from the engine, which in turn uses more fuel. Consistent and smooth acceleration can help improve fuel efficiency, while rapid and frequent acceleration can decrease it.

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