How Far East Did the Backpacker Walk?

[HT3]

Homework Statement

In reaching her destination, a backpacker walks with an average velocity of 1.34 m/s, due west. This average velocity results because she hikes for 6.44 km with an average velocity of 2.68m/s, due west, turns around, and hikes with an average velocity of 0.447 m/s, due east. How far east did she walk?

Homework Equations

Δv=Δd/Δt
Δd=(v1xt1)+(v2xt2)
Δt=t1+t2

The Attempt at a Solution

v1=2.68m/s west
t1=2402
d1=6440m west

v2=0.477m/s east
t2=?
d2=?

At this point I notice I have two variables? t2? and d2? And I do not where to continue from here.

please help me thanks in advance

 

[asd1249jf]

Well having two variable means you need two equations to solve this, right?

Your equation for average velocity turns out to be (- is west, + is east)

$$-1.34\frac{m}{s} = \frac{displacement}{t_1 + t_2}$$

Where $$t_1, t_2$$ are your independent time it takes to travel west/east from given information.

And displacement = $$d_1-d_2$$

You've managed to find $$t_1$$, which is 2388.0 sec. and $$d_1$$ is given to you. Plug that into the equation above.

You have another equation for the distance traveled east, which is

$$d_2 = 0.447 * t_2$$

Now you have two equations and two unknowns! Rest is up to your math.

 

[HT3]

Ahh I am going to have to use a system of equations i see..
thanks a lot..ill post back once i solve it

 

[asd1249jf]

Unless if you know Calculus :D There's more of a dynamics-approach of solving this.

 

[HT3]

awesome thanks so much i got the answer - this place will probably be another home for me during this physics semester :) haha thanks again man