# Motion in one dimension help?

#### Questions999

An automobile which travels with the constant speed 20 m/s passes in a section in the moment t=0s and 5 seconds later in the same section passes another automobile which travles with the speed of 30 m/s in the same direction.a) Find when the second car meets the first one.
My solution : x1=x2 20t=30(t-5) and here we find t=15 seconds.b) Whats the distance between the cars since the section until they meet? How to do this one? Just give me an idea..

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#### tiny-tim

Homework Helper
Hi Elaia06! hmm ahh! , i think you mean …​
An automobile which travels with the constant speed 20 m/s passes a line at time t=0s and 5 seconds later another automobile which travels with the speed of 30 m/s in the same direction passes the same line.
a) Find when the second car meets the first one.
My solution : x1=x2 20t=30(t-5) and here we find t=15 seconds.b) Whats the distance past the line when they meet?
x1 = 20t
x2 = 30(t-5)
x1 = x2 you already have t = 15,

so x1 = x2 = … ? #### Questions999

I did it this way, but my teacher said it was not correct Now I logged in physicsforum and I see we have the same opinion :S Even the other students had done it this way.Whatever,thank you :)

#### tiny-tim

Homework Helper
Hi Elaia06! Please remember that I rewrote the question so that it fitted your solution!

Perhaps the question means something different?

#### Questions999

So the distance question means " The two cars,since they pass the line have a distance.They keep this distance until they meet.Find the distance.This is basically what I did,found that distance.

#### tiny-tim

Homework Helper
" The two cars,since they pass the line have a distance.They keep this distance until they meet.Find the distance.
sorry, but that doesn't make any sense (how can they keep the distance if the distance is getting smaller?)

#### Questions999

Thats what it says in my book,literally.I am not english and the book I work with is translated poorly ..Whatever,thank you anyway..