Finding the instantaneous velocity on a position time graph

AI Thread Summary
To find the instantaneous velocity on a position-time graph at 0.5 seconds, a tangent line must be drawn at that point. The slope of this tangent line can be calculated using the formula (y2-y1)/(x2-x1). It is recommended to draw a longer tangent line to ensure that both the "rise" and "run" are significant, making measurements more accurate. In this case, the velocity was constant, simplifying the process to just finding the slope. The discussion highlights the importance of accurately measuring the tangent line for precise velocity calculations.
slag1928
Messages
13
Reaction score
0
I need help with a very general question. I was asked to find the instantaneous velocity of a position time graph at .5 seconds. i know to do this i need to create a line that is the tangent to that point. Here lie the problem... how on Earth do i make that line, and how do i measure the slope?



i think (y2-y1/x2-x1)for finding the slope of the tangent line? but i have no idea where that line should begin or end.

Thanks
 
Physics news on Phys.org
It doesn't matter where the line begins or ends, it still has the same slope... So I'd draw a reasonably long line so that both the "rise" (y_2-y_1) and "run" (x_2-x_1) are biggish numbers, and then calculate their ratio as you suggest. (If they're both big numbers, they're easier to measure precisely so your value for the gradient will be more accurate.)
 
thank you. turns out i was over complicating things. the velocity was constant for the points i was finding so it was as simple as finding the slope. T.T
 
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
13
Views
3K
Replies
9
Views
1K
Replies
23
Views
5K
Replies
3
Views
2K
Replies
14
Views
5K
Back
Top