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Dream Deep
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 Posted: Thu Jan 11, 2007 7:28 pm Post subject: And... More Calc |
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Clearly, this subject should be obliterated.
Oy, here it is.
In this problem, find both the net distance and the total distance traveled between times t=a and t=b by a particle moving along a line with the given velocity finction v=f(t)
v = t^2 - t - 2; a = 0, b = 3
It looks so simple, too, that's what's killing me.
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Help. Me. Please. *head/desk* |
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Incandescence
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| Posted: Thu Jan 11, 2007 9:06 pm Post subject: |
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Dream Deep:
I posted this in chat, but I'll repost. You should know the order of physics.
Acceleration=D(Velocity)/Dt
Velocity=D(Displacement)/Dt
Obviously, Displacement=Int(Velocity,a->b).
Which physically makes sense, if you care to think about it.
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Brad |
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Griffinkeeper
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| Posted: Thu Jan 11, 2007 9:16 pm Post subject: |
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All right, this is a little weird, but it works out.
The first thing I did was graph it. This looks like a shallow parabola, it actually crosses the t-axis at t=2.
So, what does this tell us?
This tells us that from 0<=t<=2, the velocity is negative, but when t>2, the velocity is positive.
What net distance means in this case is how far the particle has traveled forward, where the total distance is how much the particle has traveled period.
I know that the area under the velocity curve is equal to the distance traveled. This means that if I integrate, I'll get my answer.
So, to get the total net, this is what we do. We integrate from t=0 to t=2, and that should give us the negative distance. Then, we integrate from t=2 to t=3 to find the positive distance.
To find the total distance, add the positive distance with the absolute value of the negative distance. To find the net distance add the positive and negative distances together.
After integration, I find that the net distance is -1.5, and that the total distance was 5.16667.
To check our answer, let's take a look at the graph to see if it makes sense. Since the majority of the velocity curve is under the t-axis, (from 0 to 2) it's a sure bet that our answer is going to be negative. The positive velocity does kick in, but it doesn't have enough time to move forward.
Our answer seems to support the behavior we see on the graph.
So, where would this be useful knowledge?
It is useful for tracking the movement of an object through space, such as an artillery shell. You would want to know the total distance it travels over time, not just the vertical or horizantal distance. |
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Phorcys
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| Posted: Sat Jan 13, 2007 3:18 am Post subject: |
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Phorcys, you're right in that we could use a graph and this information to figure out the change of the velocity over time.
In this case though, the question doesn't call for it. They want to figure out the distance covered, this is represented by the space between the curve and the t-axis.
If we took the derivative of this function we would simply create a new function, that is
f'(t)=a.
Where a stands for acceleration, which is the derivative of velocity.
Incidently, f'(t)=2t-1.
This is useful for finding out how the velocity changes over time, but not so useful for figuring out the distance the particle travels.
It's easier to take the integral of f(t) from a to b. |
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Phorcys
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| Posted: Sat Jan 13, 2007 4:05 pm Post subject: |
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Oh right, I see now, I was getting confused with rates of change but this is actually something different. We are covering Integration in a few weeks.
!!! - My Maths exam in on Tuesday btw, mostly on graph work and differentiation. |
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