# Homework Help: Two Objects Approaching One Another

1. Jul 21, 2011

### Tearsandrille

1. The problem statement, all variables and given/known data
How far could you run back and forth at 6.5m/s between two friends who are 65m apart while one friend is running with a speed of 2.6m/s and the other friend is running with a speed of 1.2m/s?

2. Relevant equations
v = d/t

3. The attempt at a solution
If I establish the following:
V1 = 1.2m/s
V2 = -2.6m/s
*One of these will be negative since they are moving towards each other hence different directions.
D1 = 0.0m
D2 = 65m
*This is referring to the initial position not the distance being traveled.

Then, I input those numbers into the velocity equation.
-2.6m/s*t = d + 65m
1.2m/s*t = d

I know that the distance and time of impact between the two running towards each other will be the same. So, I can input one of the equations into the other and solve for t.

-2.6m/s*t = 1.2m/s*t + 65m
-2.6m/s*t - 1.2m/s*t = 65m
t(-3.8m/s) = 65m
t = -17.1s

This is where I am getting confused. How am I setting this up wrong to get a negative time. I don't feel comfortable with "just getting rid of the negative" under the notion time can't be negative as a few people have told me to do.

To finish the problem, I know that I can use the time calculated in the velocity equation to calculate how far I travel.

Thank you,
TS
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 21, 2011

### Aimless

What is d?

At time 0, where are the two friends? At time t, where are the two friends? Is that consistent with the above?

3. Jul 21, 2011

### cupid.callin

I couldn't understand your method properly

you need to find after how long 2 friends meet
let's assume they meet at x from origin
time t will be same for both and speeds and distances (x and ___)are given
2eqn, 2 unknown
rest i guess is easy

4. Jul 21, 2011

### Staff: Mentor

If you are using as the origin of your coordinate system the initial position of the first friend (running at +1.2 m/s), then the position of the second friend should be:

x(t) = 65m - 2.6 m/s * t

A quicker route to your time of impact is to consider the closing velocity of the two friends, as opposed to their individual velocities and positions. How fast are they closing?

5. Jul 21, 2011

### cupid.callin

Are you referring to relative velocity ?

6. Jul 21, 2011

### Tearsandrille

I see. The d I was using in the equation is the displacement, but I was treating it as position. If I treat it like position then I should get:
x(t) = 65m - 2.6 m/s * t
x(t) = 0m + 1.2 m/s * t

Since, the position, x(t), of impact will be the same I can set them equal to each other and solve for t. I can use that t for myself (running with a velocity of 6.5 m/s).

Thank you,
TS

Last edited: Jul 21, 2011
7. Jul 21, 2011

### Staff: Mentor

Yes. In this case they are running towards each other, so they are "closing", and we can refer to "closing speed" or "closing velocity".

8. Jul 21, 2011

### Tearsandrille

I believe that to find how fast they are closing 3.8m/s, which is I believe the relative velocity of one friend to the other. Is this what you were referring to? It does seem to be quicker in that I am skipping the step of setting them equal to each other.

TS

9. Jul 21, 2011

### Staff: Mentor

Yes. Saves time.