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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?
JasonRox
Sep7-04, 10:04 PM
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.
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?
JasonRox
Sep7-04, 10:24 PM
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.
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?
JasonRox
Sep7-04, 10:42 PM
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.
JasonRox
Sep7-04, 10:45 PM
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.
so distance equal to 2.1?
Let's start with runner a.
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?
More advanced:
Can you write an equation for the position of runner a as a function of time?
JasonRox
Sep8-04, 09:32 AM
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.
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