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Chase problems!

  • Thread starter napoodo
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  • #1
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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?
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi napoodo! Welcome to PF! :smile:
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! :wink:

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! :smile:

(if you're still having difficulty, show us what you've done so far, so that we know how to help)
 
  • #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
 
Last edited:
  • #4
tiny-tim
Science Advisor
Homework Helper
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4t = 10/t
4t(square) = 10/4
(square root) t (square) = (squareroot) 6
t = 2.5s?
Hi napoodo! :smile:

I'm sorry, I don't understand any of that. :confused:

You really do need to write out equations beginning with "x ="

What is the equation for the man? :smile:

(going to be now :zzz:)
 

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