Basic Kinematics Problem -- Backpacker's average velocity and distance....

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physlexic
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Homework Statement


Problem reads:
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 2.68 m/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


Vavg = delta X/ delta t

The Attempt at a Solution


I let
vw= average velocity from hiking west
xw = west displacement
tw = the time it took to travel in the west direction

ve = average velocity from hiking east
xe = east displacement
te = the time it took to travel in the east directionvw = 2.68 m/s
xw = 6.44 km = 6440 m
tw = 6440 m / 2.68 m/s = 2403 s

ve = 0.447 m/s
xe = ?
te = ?

using the avg. velocity equation as ve = xe / te I solve for xe where ve = 0.447 m/s
xe = te * 0.447 m/stherefore according to the equation Vavg = xe-xw / te-tw
However, I'm looking at other solutions and it shows the equation to be Vavg = xw + xe / te + te
I thought average velocity is a change in displacement over a change in time? So why are they adding the displacement & time as opposed to subtracting? So confused...
 
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on Phys.org
physlexic said:
I thought average velocity is a change in displacement over a change in time? So why are they adding the time as opposed to subtracting?
Because the total change in time is the sum of the two times and not the difference as the times are defined:
physlexic said:
t1 = the time it took to travel in the west direction
physlexic said:
t2 = the time it took to travel in the east direction
 
But every Average Velocity equation is x2-x1 / t2-t1
at least that's what my book defines it...the change in time is the final time minus the initial time.

Same applies for displacement but yet they are adding it here as well.
 
You need to look beyond symbols and realize what they are representing rather than simply try to apply a formula.

In this equation:
physlexic said:
But every Average Velocity equation is x2-x1 / t2-t1
t1 and t2 are the initial and final times, respectively.

In the equation in your OP, the times given are not the start and end times, it is the travel times.