## 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?

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Hi napoodo! Welcome to PF!
 Quote by napoodo 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?
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 t

x = something else t
Then 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)

 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

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## chase problems!

 Quote by napoodo 4t = 10/t 4t(square) = 10/4 (square root) t (square) = (squareroot) 6 t = 2.5s?
Hi napoodo!

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 )