A difficult 1D kinematics problem dealing with displacement and velocity

AI Thread Summary
The discussion revolves around a kinematics problem involving a backpacker's journey, where she walks with different average velocities in opposite directions. The average velocity of 1.31 m/s due west is derived from her total distance and time traveled. Calculations reveal that she walked 829.87 meters east after hiking 6.44 km west. The confusion arises between total displacement and the distance walked east, clarifying that the question specifically asks for the latter. Ultimately, the correct answer for the distance walked east is 829 meters.
kathypoot
Messages
2
Reaction score
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.
 
Physics news on Phys.org
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?
 
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... :(
 
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top