- #1
vkrock
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Homework Statement
In reaching her destination, a backpacker walks with an average velocity of 1.32m/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 .447m/s due east. How far east did she walk?
Homework Equations
Unsure. However,
V^2 = V(initial)^2 + 2ax seems like it would yield an answer if you treat walking in the opposite direction like slowing down (as far as average velocity is concerned).
The Attempt at a Solution
I tried using the equation mentioned above, thinking that turning around and walking east at .447m/s was like having an acceleration of -.447m/s because her positive average velocity due west is being decreased as she walks east. Unfortunately doing so yields answers of 24.34m which seems unreasonably low.
The other method I tried which yielded results was guess and check, using that she walked 6440m west at 2.68m/s which means for a total of 2403 seconds. I tried out a few distances east she could walk and checked to see what kind of impact that would have on her average velocity, this was the best match i could find:
6440m - 805m / 2403s + (805m/.447m/s) = 1.34m/s
i picked 805meters east, so i subtracted that from the 6440m west since it's like she's "unwalking" that distance, for time I add the 2403seconds it took her to get to 6440m west and the time it would take her to walk those 805m east which I got by dividing the 805m by her easterly velocitiy to come up with the time it would take her to achieve that distance.
I am wondering if the second method of guess and check and the answer it yielded seems reasonable. I am also curious if there is an easier way to do it but I was unable to get anything significant using the equations provided thus far in the course (which haven't been many).