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The acceleration 'component' is then $–9.8 \text{m s}^{-2},$ but, rather sloppily, we often say simply that (having chosen the upward direction), the stone's acceleration is negative! Note that the stone's upward acceleration component is always $–9.8 \text{m s}^{-2},$ , whether the stone is on its way up and slowing down, or on its way ...
The magnitude of the acceleration vector along the path is the time rate of change of speed. The magnitude of the acceleration vector normal to the path is the centripetal acceleration as it goes around the instantaneous radius of curvature ρ(t) ρ (t). The combined magnitude is the combination of the above and does not have a direct ...
Assume constant acceleration and find the acceleration in terms of g during a) the launch and b) the speed reduction. The basic strategy to find acceleration I am using is to calculate two velocity equations: one between (0 m/s, 0 m) and (1.6 m/s, 5.0 micrometers); the second between (1.6 m/s, 5.0 micrometers) and (0 m/s, 1.0 mm).
Now, to measure the average acceleration, we do the same thing as above: $\text{time} * \text{acceleration} = \text{velocity} \implies \text{acceleration} = \frac{\text{velocity}}{\text{time}}$. Now just look at the units: On the right side, you already have meters per second for speed, and now you are looking at the change of this speed over ...
Zero acceleration is zero. It's different from constant acceleration as the latter means any value >0 or <0 but not equal to 0. From the graph, if there would have been constant acceleration, the graph would have been different. So, although zero is a constant, zero acceleration is zero acceleration, not constant acceleration.
If velocity and acceleration have the same sign, you speed up. If velocity and acceleration have opposite signs, you slow down. Negative velocity = You move backwards. Negative acceleration = You slow down or you go faster in the backwards direction. Let's say you have this simple coordinate system:
0. Yes you are correct, Acceleration is the rate of change of velocity, in short if you are accelerating the more the time passes the more is your velocity and the distance travelled. The m/s2 of acceleration is actually meter per second per second (m/s / s) which is the rate of change of velocity so summing up the indices it is written as m/s^2.
I don't think that works at all describing the weak nuclear force, but works well for classical mechanics. Does acceleration appply regarding WNF? Newton's second law is: A force causes an acceleration inversely proportional to the object's mass. Apply the same force to a different mass, you get a different acceleration that still satisfies F=ma.
The difference is that acceleration is absolute whereas velocity is relative. In other words, the local laws of physics are exactly the same for two objects moving with a constant velocity relative to one another, so there is no local experiment that can determine whether one is moving and the other is stationary.
Go back and refit the points with the equation I gave you, fitting for 'a' and 'v', then 'a' is simply the acceleration. Using the linear slope in this case is useless, outside a rough guess. To properly solve for acceleration, first you need to find out the velocity as a function of time, then you can find the acceleration.