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Position and acceleration graphs
- Thread starter dlee9
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- Acceleration Graphs Position
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To determine position and acceleration graphs from given velocity graphs, it's essential to understand that acceleration is represented by the slope of the velocity graph. The acceleration graph cannot be option F, as it does not show any abrupt changes in slope. Additionally, the position graph cannot be option I, since a negative slope at t=7 indicates that the velocity should also be negative. Correctly identifying these relationships is crucial for accurate graph interpretation. Understanding these concepts will help clarify the connections between the different types of motion graphs.
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CWatters
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I think you should check the first and last answers..
The graph for ax cannot be F. Acceleration is the slope of the velocity curve. At no point does the slope of the velocity curve change abruptly.
The graph for the position x cannot be I. At t=7 the slope of the position graph I is negative so the velocity should be negative.
The graph for ax cannot be F. Acceleration is the slope of the velocity curve. At no point does the slope of the velocity curve change abruptly.
The graph for the position x cannot be I. At t=7 the slope of the position graph I is negative so the velocity should be negative.
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...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
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...
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