Solving for When and Where Two Movements Meet

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The discussion revolves around solving a physics problem involving a man racing and a woman biking, focusing on determining when and where they meet. The man starts 10 meters ahead of the finish line at a constant speed of 10 m/s, while the woman starts from rest and accelerates at 4 m/s². Participants suggest writing equations for both movements to find a common point where their distances equal each other. The user expresses confusion about setting up the equations correctly, particularly regarding the use of velocity and acceleration. Clarification on the need for consistent equations is emphasized to solve the problem effectively.
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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?
 
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Welcome to PF!

Hi napoodo! Welcome to PF! :smile:
napoodo said:
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)
 
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:
napoodo said:
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:)​
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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