How Do You Find Instantaneous Velocity at 1s?

AI Thread Summary
To find the instantaneous velocity at 1 second for an object moving along the x-axis, the position versus time graph is analyzed. The initial position is given as -2 m, and the equation of the line derived from points (0, -2) and (2, 7) is y = (7/2)x - 2. The slope, representing the average velocity over the interval from 0 to 2 seconds, is calculated as (y2 - y1)/(x2 - x1). However, the instantaneous velocity at a specific point requires the correct interpretation of the slope at that point, which may differ from the average velocity. The discussion highlights the importance of verifying calculations and understanding the distinction between average and instantaneous velocity.
hackett5
Messages
9
Reaction score
0
The position versus time for a certain object moving along the x-axis is shown. The object’s initial position is −2 m. Find the instantaneous velocity at 1s.
http://img23.imageshack.us/img23/5285/phy1w.jpg

Using points (0,-2) and (2,7) I found the equation of the line to be =7/2x-2. Since this is a straight line I thought the tangent line would have the same slope. So, at 1s the instantaneous velocity would be 7/2, but this is not the correct answer. Where am I going wrong?
 
Last edited by a moderator:
Physics news on Phys.org
Slope of the line = (y2 - y1)/(x2 - x1).
From 0 to 2 seconds this will be the velocity of the particle.
 
Thanks for pointing out my now obvious mistake! I didn't even think to check to see if I had gotten the slope correct, I just assumed my understanding of instantaneous velocity was wrong.
 
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...
Back
Top