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chase problems! |
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| May26-08, 06:50 PM | #1 |
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chase problems!
omg i have so much trouble doing these problems... can anyone help? and show process in which you took? thank you so much!
a man starts 10m ahead of a finish line and races at a velocity of 10m/s. A women rides a bike from 0m (starting line) and accelerates 4m/s(square). when and where do they meet? |
| May26-08, 07:03 PM | #2 |
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Hi napoodo! Welcome to PF!
![]() Isn't that typical of a man … starting at the finish line! ![]() I assume you mean that they started at the same time? Just write an equation in x and t for each of them: x = something tThen they meet when their x is the same, so you just write something t = something else t, which is an equation only in t, which you can solve! ![]() (if you're still having difficulty, show us what you've done so far, so that we know how to help) |
| May26-08, 07:11 PM | #3 |
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ok well i tried, and this is what i got. [Of woman]==> at=x=d/t <===[Of man] aka a=v/t and v = d/t and i put v is the same...
4t = 10/t 4t(square) = 10/4 (square root) t (square) = (squareroot) 6 t = 2.5s? would that time be right? or do you need the same equation to make it equal each other O_O because one side uses velocity, and time, and the other side uses acceleration and time shouldn't d(distance) be the same? = both |
| May26-08, 08:07 PM | #4 |
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chase problems!![]() I'm sorry, I don't understand any of that. ![]() You really do need to write out equations beginning with "x =" What is the equation for the man? ![]() (going to be now |
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