How Do You Calculate Velocity from an Acceleration-Time Curve?

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To calculate velocity from an acceleration-time curve, the discussion focuses on two time intervals: 0 < t < 10 seconds and 10 < t < 26 seconds. For the first interval, the correct velocity equation is v = 5t, while for the second interval, the initial velocity must be determined based on the previous segment. The challenge arises in correctly applying the initial velocity for the second interval, which can be derived from the first interval's result. Clarification is needed on whether to use the initial velocity from the entire motion or just the segment being analyzed. Understanding these nuances is crucial for accurately formulating the velocity equations for each time segment.
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1. We are given a simple acceleration vs time curve. It shows that the object has a constant acceleration of 5 m/s2 for 10 seconds, and then from 10 to 26 seconds has a constant acceleration of 15 m/s2. The question asks me to give an equation for velocity as a function time using only the variable T.



2. V = V initial + a*t
?




3. This is a multi part question, and I have gotten the first part correct. The first part simply asks for the equation from 0 < t <10. I responded that v = 5*t , and got it correct. A similar question within the same problem asks, "what if the initial velocity is 35 m/s" I responded v = 35 + 5*t and got it correct.

The next part is to give an equation (using the same variables) but for 10 < t < 26. I have tried v= v initial +at to no avail.


The real problem is, because this is online homework, I can't tell if I am doing it wrong, or typing it wrong. I've looked through the book and my notes, and have not seen anything to help me in the right direction.

So what I am doing wrong for the 10 < t < 26 portion?

Thanks to all.
 
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Audaciter said:
... "what if the initial velocity is 35 m/s" I responded v = 35 + 5*t and got it correct.

The next part is to give an equation (using the same variables) but for 10 < t < 26. I have tried v= v initial +at to no avail.[/b]
...

So what I am doing wrong for the 10 < t < 26 portion?

Thanks to all.

What does vinitial correspond to for this part. It might be better to call it v10.

When 10 < t < 26, for what amount of time has the object has been accelerating at 15m/s2?
 
Thank you for the quick reply.

I would assume that V inital for 10 < t 26 is simply 5*t, from the first portion of the question, is that correct? Meaning v = 5*t + 15t.

The object instantaneously accelerates from 5 m/s2 to 15 at t = 10. The graph looks like a single step if you know what I mean.

It has the same basic shape as this graph, but mine is a "step up" rather than a "step down" with different numbers obviously.

[PLAIN]http://img268.imageshack.us/img268/8762/acvt9nz.jpg
 
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Audaciter said:
1. We are given a simple acceleration vs time curve. It shows that the object has a constant acceleration of 5 m/s2 for 10 seconds, and then from 10 to 26 seconds has a constant acceleration of 15 m/s2. The question asks me to give an equation for velocity as a function time using only the variable T.

It sounds like you're going to need to employ the unit step function in your expression.

edit: Unless of course you only need to express the time interval from 10 to 26 seconds. Does the problem text make it clear whether or not the initial 35m/s velocity mentioned elsewhere is to be assumed in this section too?
 
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Sorry, I didn't make it clear. There are 2 questions dealing with the same graph, and each of those has two parts.

Question 1 assumes a initial velocity at t = 0 is v = 0.
Then 1a) asks for what I discussed above. Equation for velocity as a function of time for t <10. 1b) asks for the function for 10 < t < 26.

Question 2 then assumes an initial velocity at t = 0 is v = 35; 2a and 2b are the same as question 1.

So, if you are only doing the time interval for 10 < t < 26 (with either question) do you use the initial velocity of THAT segment of the motion, or the initial velocity of the entire motion?

I guess my root problem is I don't know how to get an equation for the second time interval. I have correctly gotten the equation for 1a and 2a, but can't seem to get 1b and 2b...I hope I am not confusing this more.

Thank you.
 
If you only need to produce an equation for a specific time interval, then I'd use the velocity at the beginning of the interval as the initial velocity and forget the past.
 
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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