A difficult 1D kinematics problem dealing with displacement and velocity

In summary, the backpacker walks with an average velocity of 1.31 m/s, due west, for a total distance of 6.44 km. After turning around, she walks with an average velocity of 0.475 m/s, due east, for a total time of 2535.4 seconds. Therefore, she walked 829.87 meters (or 0.829 km) east.
  • #1
kathypoot
2
0
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.
 
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  • #2
kathypoot said:
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.
Welcome to Physics Forums.

There is sufficient information to solve the problem.

HINT: What is the definition of average velocity?
 
  • #3
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
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.
 

Related to A difficult 1D kinematics problem dealing with displacement and velocity

1. What is 1D kinematics?

1D kinematics is the study of motion in one dimension, where the only factors considered are position, time, and velocity. This type of motion is often represented on a straight line with a positive or negative direction.

2. What does displacement refer to in 1D kinematics?

Displacement is the distance and direction of an object's position from its starting point. It is a vector quantity and is typically represented by the symbol "Δx". In 1D kinematics, displacement is measured along a straight line.

3. How is velocity calculated in 1D kinematics?

Velocity is the rate of change of an object's displacement with respect to time. In 1D kinematics, it is calculated by dividing the change in position (displacement) by the change in time. This can be represented by the equation v = Δx / Δt.

4. What is a difficult 1D kinematics problem?

A difficult 1D kinematics problem typically involves multiple variables, such as initial and final positions, time, velocity, and acceleration. It may also require the use of kinematic equations and mathematical manipulation to solve.

5. What strategies can be used to solve a difficult 1D kinematics problem?

Some strategies that can be used to solve a difficult 1D kinematics problem include identifying known and unknown variables, setting up a clear diagram or coordinate system, using the appropriate kinematic equations, and checking for units and significant figures in calculations.

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