# Constant Velocity

## 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.

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 cant 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