# At what time will runner B catch up with runner A?

1. Sep 7, 2004

### JohnQ

Runner A, who runs with an average speed of 3.0 m/s, starts out at 3:00 P.M. Runner B, who runs with an average speed of 4.0 m/s, starts after A from the same place exactly 5 min later.
a.) At what time will runner B catch up with runner A?
b.) If the runners stop when B catches A, how far do they run?

So far I have 4.0 m/s*300 s = 1.2 Km and 3.0 m/s x 300s= 0.9 km, but I don't even know if this is the right aproach, can anyone help me from here?

2. Sep 7, 2004

### JasonRox

They are two linear equations(with one solution I believe), find the point of intersection.

I will not tell you the answer, but merely try to open your eyes.

You know the velocities of A and B.

What formula do you use to find the velocity?

Try to combine the formulas to come up with an answer.

Hint: You need to use algebra. The distance travelled must be equal for both because that is when B catches up to A.

Show me some more work, and I'll be back to give more advice.

Note: I can tell you more right now, like most people, but doing it on your feels so much better.

3. Sep 7, 2004

### JohnQ

Trying

so....
If I let the time A runs be "t", then the time B runs is "t- 5".
...4.0 m/s= t-5
and 3.0= t

So... 4.0 m/s= 3.0-5
I'm I on the right track?

4. Sep 7, 2004

### JasonRox

Huh?

To take that route I believe you will need to convert the minutes into seconds since your velocity is 4.0m/s.

Look at what you wrote. You said 4=3-5. We know that's incorrect already.

The formula for velocity is $$v=d/t$$.

Because there is two different velocities, there is two different equations to work with.

With the two equations that you get, combine them.

Note: Remember, d will be the same for both.

5. Sep 7, 2004

### JohnQ

Still trying

so...for B v=1.2 km/ 3.0

and for A v= .9 km/3.0, IF RIGHT, WHAT UNITS IS 3.0 IN, HR?

6. Sep 7, 2004

### JasonRox

Wrong. I'm having a hard time trying to understand where you got this.

I'll give you the equation for B, since you kind of got it in your second post. A is very similiar.

$$4.0m/s = \frac{d}{t_1-300s}$$ , 300s = 5min (converted into seconds).

You had t-5, but that was incorrect. Find equation A. Remember d is the same for both.

Combine both equations by isolating a term(d or t_1).

Note: I must go bang my head trying to fully comprehend SR.

Hopefully someone takes over if you are still stuck.

7. Sep 7, 2004

### JasonRox

Also, read the first 2 chapters of your textbook. It will talk about dimensionalities(metres, seconds) and most likely average velocities.

That would help a lot.

There is also a forum for homework help, and they have helpful people there. Believe it or not, they have regulars for helping out.

Please avoid homework related problems in this forum. I made the mistake myself, so it's not big deal, so I'm letting you know.

8. Sep 7, 2004

### JohnQ

?

so distance equal to 2.1?

9. Sep 8, 2004

### pervect

Staff Emeritus

If runner a moves at a velocity of 3 m/s, and he runs for 1 second, how many meters does he move?

If he runs for 2 seconds, how many meters does he move?

Can you write an equation for the position of runner a as a function of time?

10. Sep 8, 2004

### JasonRox

That's what I was saying. I even gave the equation itself.

I would love to show you how it's done, but it is better to work it out yourself. Once you get that breakthrough, everything else becomes a little easier.

Note: The answer might be 2.1m because I never checked.