Velocity squared vs position graph

AI Thread Summary
The discussion centers on the interpretation of the slope from a velocity squared versus position graph. The calculated slope, with units of cm/seconds squared, is questioned regarding its relation to acceleration, which is expected to be 9.80 m/sec². Clarification is provided that the slope represents twice the acceleration (2a) when using the equation v² = u² + 2as, where 's' refers to displacement, not time. Confusion arises from the terminology, as 's' is often misinterpreted as seconds. Ultimately, the slope does help determine acceleration when correctly applied in the context of the graph.
azn4lyf89
Messages
17
Reaction score
0
I calculated a slope from a best fit line of a velocity squared vs position graph but I don't understand what this slope represents. My units come out to be in cm/seconds squared, so does that mean that's my acceleration? It seems to be way off from 9.80m/sec^2 which should be my acceleration. Or does this slope help me determine to acceleration?
 
Physics news on Phys.org
Well from v^2=u^2+2as

If you plot v2 vs. s, your gradient should be 2a
 
rock.freak667 said:
Well from v^2=u^2+2as

If you plot v2 vs. s, your gradient should be 2a

Isn't that only if the graph was velocity squared vs time? Because s is seconds right?
 
azn4lyf89 said:
Isn't that only if the graph was velocity squared vs time? Because s is seconds right?

s is displacement.

t is usually used to inidicate time.
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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

Back
Top