Calculating Relative Velocity for Walking on a Treadmill

AI Thread Summary
The discussion focuses on calculating relative velocity and displacement while walking on a treadmill moving at -4 km/h. The velocity relative to the treadmill is determined to be +4 km/h, as one must walk at this speed to remain stationary on the treadmill. The velocity relative to the floor is concluded to be 0 km/h, indicating no net movement. Total displacement relative to the treadmill after 15 minutes is calculated to be +1 km, while displacement relative to the floor is 0 km. The participants express some confusion but engage in the problem-solving process.
Jacobpm64
Messages
235
Reaction score
0
You exercise by walking for 15 minutes on a treadmill that has a level track that moves at a velocity of -4km/h.

What is your velocity relative to the treadmill?

What is your velocity relative to the floor?

What is your total displacement relative to the treadmill when you finish exercising?

What is your total displacement relative to the floor when you finish exercising?
 
Physics news on Phys.org
At least make some guesses of your own. Can you show us that you have put some thought into the problem?
 
i can try..

What is your velocity relative to the treadmill?
I think this one would be +4km/h. Hard for me to explain why i think this, but I'll try. Uhmm the treadmill is moving backwards.. and you have to run forwards at the same speed as the treadmill so that you won't fly off the treadmill in either direction (run too slow.. and you fall off the back end.. run too fast and you fall off the front).. So, if you run at the same speed as the treadmill, but in an opposite direction, you will end up staying in one place (which is the object of the treadmill)... So, your velocity relative to the treadmill would be +4km/h? I don't know.. but that's what i think


What is your velocity relative to the floor?
Since you aren't actually moving anywhere, the velocity relative to the floor is 0km/h? I think..

What is your total displacement relative to the treadmill when you finish exercising?
Because I said the velocity that i move relative to the treadmill is +4 km/h, then to find the displacement, I can solve for displacement in the velocity formula (v = d/t), to get d = vt. Plugging in the numbers, I get 15 min x 4km/h.. convert hours to minutes... 15min x 1/15 km/min = +1 km
So, my answer is +1 km.

What is your total displacement relative to the floor when you finish exercising?
Same approach, using the answer from the 2nd question this time... d = vt... So, plugging in numbers, 15min x 0km/min = 0km
So, my answer is 0km.

Are these right? Or? Because, I do feel confused.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
17
Views
1K
Replies
2
Views
1K
Replies
56
Views
3K
Replies
38
Views
3K
Replies
3
Views
3K
Replies
5
Views
2K
Replies
35
Views
5K
Back
Top