Solve Constant Velocity: Dogs Collision in 85m

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Homework Help Overview

The problem involves two dogs running towards each other from a distance of 85 meters, with Dog A moving at a velocity of 3.5 m/s and Dog B at -2.8 m/s. The objective is to determine the time it takes for the dogs to collide.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the equation of motion but expresses confusion about the setup and solution process. Some participants suggest writing out the equations of motion for both dogs and setting their position functions equal to find the collision time. Others question the clarity of the explanation and seek simpler terms for understanding.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the formulation of position functions, but there is no explicit consensus on the approach or understanding of the concepts involved.

Contextual Notes

The original poster expresses difficulty in grasping the problem, indicating a need for clearer explanations. There is also a mention of the dogs' starting positions and the implications for their respective equations of motion.

baileya
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Homework Statement



Two dogs stand facing each other 85 meters apart. They begin to run towards each other. Dog A runs at a constant velocity of 3.5 m/s; Dog B is going the opposite direction at a constant -2.8 m/s. How long before the dogs collide?

Homework Equations



Xfinal-Xinitial = vt

The Attempt at a Solution



I looked through my notes from class and tried to work through it. I also googled it but couldn't figure it out.
 
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Write out your two equations of motion.

For dog 1 you would simply have x(t)=3.5t if you choose him to be at zero on your coordinate axis.

For dog 2 what would you have?

Once you set those two position functions equal to each other you will be able to solve for the time. Then simply plug that time into either of the two functions and they will provide the same result of where the dogs collide.
 
haha this probably makes me sound really stupid but you're going to have to explain it in more simple terms... if that's possible.

I just really don't get this question.
 
Okay, so you have learned that position i will call it 'x' is equal to velocity multiplied by time or x=vt for motion in 1-D. What this really means is that given some velocity
'v' you substitute in any value for time 't' (greater than zero since time can't be negative) and get the position of whatever object at that chosen time. So it really means that position is your dependent variable and time is your independent variable...or position is a function of time written as x(t).

The second position function is a little trickier than just x2=-2.8t because it doesn't start at x=0 it starts at x=85m (since i choose the first dog to be at zero and the dogs are 85m apart). If you were to sub zero in for time in the equation above you would get that x2=0 which is incorrect. So you need to add a constant that when time equals to zero makes x2=85...

Pretty much all you are dealing with is functions just like what was covered in algebra when you discussed lines often given the general formula y=mx+b
 

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