# A difficult 1D kinematics problem dealing with displacement and velocity

1. Sep 9, 2009

### kathypoot

In reaching her destination, a backpacker walks with an average velocity of 1.31 m/s, due west. This average velocity results because she hikes for 6.44 km with an average velocity of 2.54 m/s, due west, turns around, and hikes with an average velocity of 0.475 m/s, due east. How far east did she walk?

Is there not enough information? I feel like they should have given me another variable or something.

I figured out that it took her 2535.433 seconds to hike for the 6.55km stretch. But I'm really in the dark on this one, I appologize. I've looked all over the internet to try and find a problem simular to this one but I had no luck.

2. Sep 9, 2009

### Hootenanny

Staff Emeritus
Welcome to Physics Forums.

There is sufficient information to solve the problem.

HINT: What is the definition of average velocity?

3. Sep 9, 2009

### kathypoot

Distance/time = average velocity

D/T = 1.31

t1 = time walking west
t2 = time walking east
d1 = distance walked west = 6.44 km = 6440 meters
d2 = distance walked east

T = t1 + t2
D = d1 - d2

t1 = distance/speed = 6440/2.54 = 2535.4 seconds

t2 = d2/v2 = d2/0.475
so d2 = 0.475*t2

Now use all this in the equation for D/T.
D/T = (d1 - d2)/(t1 + t2) = 1.31
(6440 - 0.475*t2)/(2535.4 + t2) = 1.31
6440 - 0.475*t2 = 3321.42 + 1.31*t2
1.786*t2 = 3118.58
t2 = 1747.11 seconds

d2 = t2*v2 = 829.87

Check by calculating the displacement and time and then seeing if we get the average velocity we started with.
displacement = d1 - d2 = 5610.13 meters
time = t1 + t2 = 4282.54 seconds
average speed = displacement/time = 5610.13/4282.54 = 1.31 m/s

So she walked 5610.13 meters (or 5.61 km) east

but my answer is still wrong... :(

4. Sep 9, 2009

### Jebus_Chris

What you did is correct. The question is asking for how far east she traveled. That would be d2, 829, not 5610 which is the total displacement which 5610m west.