Solving for When and Where Two Movements Meet

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Homework Help Overview

The problem involves two individuals: a man racing at a constant velocity and a woman accelerating from rest. The man starts 10 meters ahead of a finish line, while the woman starts from the starting line. The objective is to determine when and where they meet during their respective movements.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss formulating equations for both the man and the woman based on their respective motions. There is an attempt to equate their positions to find the meeting point. Questions arise regarding the correct setup of equations and the interpretation of distance and time in the context of their movements.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the equations needed to solve the problem. Some guidance has been offered regarding the formulation of equations, but there is no consensus on the correct approach or solution yet.

Contextual Notes

Participants are navigating the complexities of using different motion equations, with one involving constant velocity and the other involving acceleration. There is uncertainty about how to equate these differing forms of motion.

napoodo
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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:)
 

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