Finding Instantaneous Acceleration/Velocity

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To find instantaneous velocity from an acceleration vs. time graph, locate the tangent at the desired point, while for a position vs. time graph, the instantaneous velocity is determined by taking the derivative of the function. The area under the curve of a position vs. time graph represents velocity over an interval, not at a specific point. For a more efficient method, taking the derivative of the position function and the integral of the acceleration function can yield quicker results. This approach can enhance accuracy and reduce the time spent on calculations. Understanding these methods is crucial for solving related physics problems effectively.
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Hi, you guys seem really helpful so I didnt think the template was necessary as my question dosent really fit those guidelines.

Im currently working on a review sheet for my Physics test tomorrow, and am having trouble finding the instantaneous velocity from looking at a acceleration vs. time graph and the instantaneous velocity from a position vs. time graph.

the problems i need help with are 1c and 2d on this page:http://www.jburroughs.org/science/mschober/consta/sframe.htm

Im just looking for a general method on how to complete these problems.

thanks so much in advance, Adam
 
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using an acceleration vs. time graph you simply need to find the tangent at that point. remember Lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}

And for velocity from a position vs. time graph you just need to find the area under the graph. \sum f(x) \Delta x
 
djeitnstine said:
using an acceleration vs. time graph you simply need to find the tangent at that point. remember Lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}

And for velocity from a position vs. time graph you just need to find the area under the graph. \sum f(x) \Delta x

thanks for the reply, but I am still kind of confused. When looking at a position vs. time graph, wouldn't the area under the curve be the velocity over that time interval, not the specific time?
 
Oops I have all of that backwards.

let me rewrite that:

using a position vs. time graph you simply need to find the tangent at that point. remember Lim _{\Delta x \rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}

And for velocity from a acceleration vs. time graph you just need to find the area under the graph. \sum f(x) \Delta x

I'm really sorry if I confused you.
 
ok gotcha. Just curious, is there any other simpler way to find the inst. velocity that finding the tangent? It takes a long time and seems to be pretty inaccurate.
 
Just take the derivative of the function at hand for the position/time graph and take the integral of the acceleration/time graph (derived using the formulas I gave you above.)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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