# Simple but confusing

clandarkfire
Posers and Puzzles 25 Sep '10 00:32
1. 28 Sep '10 20:32
Originally posted by Thomaster
Hi, I'm Kareem. I'm 16.

Yup.
I can't believe that you and clarkandfire have the same name and are both 16
years old. ðŸ˜›

What are the odds on that?
2. clandarkfire
Grammar Nazi
28 Sep '10 22:05
Originally posted by iamatiger
Are we doing your maths homework Clandfire?

Physics actually.
3. 29 Sep '10 07:041 edit
I can't believe that you and clarkandfire have the same name and are both 16
years old. ðŸ˜›

What are the odds on that?
100%
We are BFFs.
4. 04 Oct '10 04:06
Originally posted by wolfgang59
Regardless of their accelerations the current fastest car will pull ahead.
I'm inclined to agree with you: Car A is going 60; Car B is going 40. Seems pretty simple all right.

As to the police question: It was stated that the car going 120 never alters his speed so why would he be going 240 when caught? The posted answer seems to be x2. With a constant acceleration of 10 kph/s it will take 12 seconds to reach 120 kph.
5. joe shmo
Strange Egg
04 Oct '10 05:51
Originally posted by Tatanka Yotanka
I'm inclined to agree with you: Car A is going 60; Car B is going 40. Seems pretty simple all right.

As to the police question: It was stated that the car going 120 never alters his speed so why would he be going 240 when caught? The posted answer seems to be x2. With a constant acceleration of 10 kph/s it will take 12 seconds to reach 120 kph.
Thats the police officers velocity when they have traveled equal distances. If the police officer speed is constant after v=120 he will never catch the speeder.
6. uzless
The So Fist
04 Oct '10 06:58
Originally posted by joe shmo
Thats the police officers velocity when they have traveled equal distances. If the police officer speed is constant after v=120 he will never catch the speeder.
You guys make this stuff way too hard to explain. Much simpler answer:

It takes 12 seconds to match the cars speed if you are accelerating at 10km/s. This means you have stopped the gap between you and him from increasing. In effect, you are now both motionless relative to one another.

So to catch him you need to accelerate 12 more seconds to get even with him for a total of 24 seconds....ie 240 km/h.
7. 04 Oct '10 16:102 edits
Originally posted by joe shmo
Thats the police officers velocity when they have traveled equal distances. If the police officer speed is constant after v=120 he will never catch the speeder.
That's true, and it did occur to me after going to bed and I contemplated getting up and noting that. I didn't.

But yeah, I agree with the 24 seconds/240 kph, and would like to add that they will have traveled 8.3333 km from the point where the chase began.
8. 06 Oct '10 06:31
Originally posted by clandarkfire
But as far as I can see, Joe Shmo is right.

Next:

A speeding motorist travels past a stationary police officer at a speed of 120km/hr. The officer immediately begins pursuit at a constant acceleration of 10km/hr/s.

a. How long will it take for the police officer to reach the speeder, assuming the speeder maintains a constant speed?

b. How fast will the officer be traveling at that time?
there is incomplete data here. since what is being asked involves distance, there is no way to calculate it without first knowing the distance between the two at the time the pursuit is begun.
9. 06 Oct '10 14:59
Originally posted by kyletoybits
there is incomplete data here. since what is being asked involves distance, there is no way to calculate it without first knowing the distance between the two at the time the pursuit is begun.
It says: ''The officer immediately begins pursuit (...)''
10. wolfgang59
Mr. Wolf
06 Oct '10 19:26
Originally posted by kyletoybits
there is incomplete data here. since what is being asked involves distance, there is no way to calculate it without first knowing the distance between the two at the time the pursuit is begun.
there is no reference to distance in the question.
velocity and acceleration are vector quantities which an object possesses at any given point.