# Thread: physics motion

1. ## physics motion

. Two runners approaching each other on a straight track have constant speeds of 4.50 m/s and 3.50 m/s, respectively, when they are 100 m apart. How long will it take for the runners to meet, and at what position will they meet if they maintain these speeds?

2. well you have 2 equations, the one of the guy running 4.5 /s and the one 100m away running 3.5 m/s. So when wil they meet? when the two equatinos intersect! so when does 4.5x = -3.5x + 100?
as to the second part, you cannot say where they will meet, since you are only given a distance and not a start/end point. I mean you can tell how many meters each one will run before they meet but you cannot say where.

3. Hello, cruzangyal!

Two runners approaching each other on a straight track
have constant speeds of 4.50 m/s and 3.50 m/s, respectively,
when they are 100 m apart.
How long will it take for the runners to meet,
and at what position will they meet if they maintain these speeds?
Here's a back-door approach to this problem . . .

They are approaching each other at a combine speed of $4.50 + 3.50 \:=\:8$ m/s.

It is as if one of them is stopped and the other is approaching at 8 m/s.

To cover 100 m, it will take: . $\frac{100}{8} \:=\:\boxed{12.5\text{ seconds.}}$

In 12.5 seconds, the faster runner has covered: . $4.50 \times12.5 \:=\:56.25\text{ meters}$
. . and the slower runner has covered: . $3.50 \times12.5 \:=\:43.75\text{ meters}$

4. Perhaps you could think of it this way. Since they're running toward each other, the time until they meet would be 100/(4.5+3.5)=25/2=12.5 seconds.

In that time one travels (12.5)(4.5)=56.25 m

and the other (12.5)(3.5)=43.75 m

The position they meet is like this:

Code:
                  meeting point
|
|
\|/
****************************************************
|<------------ 56.25-------->|<------43.75--------->|
EDIT: It would appear Soroban and I are on the same 'page' .