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Instantaneous Acceleration Help PLEASE ~

  • Thread starter AznBoi
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  • #1
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Instantaneous Acceleration Help PLEASE!!!~~~

I have a hw problem on Instantaneous acceleration:

The engine of a model rocket accelerates the rocket vertically upward for 2seconds as follows: At t=0, speed=0; At t=1s, s=5m/s; At t=2s, s=16m/s.

a) find the average acceleration during the 2s interval and
b) find the instantaneous acceleration at t=1.5s

My work so far:

a) avg acceleration:

a=v/t , (16m/s -0m/s)/(2s-0s)
= (16m/s)/2s

a=8m/s^2 Is that right??

b) I need help on this one:

What is the easiest and simplest way to find the instantaneous accerlation? What is the simplest equation?


I found an equation on a site but I don't understand what the variables mean and what numbers I have to put in.

Equation?: a= lim t->0 (delta)v/(delta)t

On site: http://rockpile.phys.virginia.edu/arch2.pdf#search="instantaneous acceleration"

Don't understand what it means and what numbers I have to subsitute!!

BTW THANKS A LOT!!! =D
 
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Answers and Replies

  • #2
berkeman
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In your previous thread, you mentioned that this is from pre-calc physics, so I'll have to make some assumptions here. If you had calculus as part of this class, you would fit a smooth curve to the 3 data points, and differentiate that curve to get the instantaneous acceleration. Then use an integration to get the average of the instantaneous acceleration.

But since this is pre-calc, I'll assume that you are only supposed to use the discrete data points (this may be wrong, however -- maybe you're supposed to interpolate a few extra points for example). So the average acceleration is the average of the 3 data points you are given (not just the first and last).

To find the discrete instantaneous acceleration, you would take a difference between adjacent velocity data points, and divide by the time difference. Part of the problem with being given so little data, is that you don't know whether to take the left-right delta or the right-left delta. I'd use the right-left delta here, since the acceleration at 0 is probably the least. So I'd get something like:

t=0 -- accel = 0
t=1 -- accel = 5-0 = 5m/s^2
t=2 -- accel = 16-5 = 11m/s^2

Then the average acceleration would be the average of those 3 values of instantaneous acceleration. Since this is a discrete calculation, it is only an approximation of the real average acceleration. In fact, you can bound the average acceleration between the left-right and right-left discrete calculations. Do you see how you would do that?


EDIT -- fixed a typo....16-5=11, not 9. Doh!
 
  • #3
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So you subtract the speeds of two points?? Is learning calculus more beneficial when it comes to Physics?? You can solve problems in physics without calculus right? with equations?

So 16-5= 11m/s^2 is that an exact number?

The answer is about 12m/s^2 on the answer key. I just don't know how they got that.
Isn't there an equation using:

dv/dt or something??


Thanks.
 
  • #4
berkeman
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AznBoi said:
Is learning calculus more beneficial when it comes to Physics?? You can solve problems in physics without calculus right? with equations?
Here's a recent thread discussing calculus & physics:

https://www.physicsforums.com/showthread.php?t=130461

AznBoi said:
So 16-5= 11m/s^2 is that an exact number?

The answer is about 12m/s^2 on the answer key. I just don't know how they got that.
Isn't there an equation using:

dv/dt or something??
Well, interpolating two extra points at 0.5s and 1.5s, and taking the lower bound sum and upper bound sum, I get between 4.5m/s^2 and 8.5m/s^2. So the book must be doing the simple discrete calculation, with the max right-left deltas. They are averaging 5 and 16 over 2s, and getting 21/2 m/s^2 ?


EDIT -- this is a weird problem. I don't think I'll be able to be of much more help. You may just have to wait for class to get clarification. Or maybe another homework helper will stop by and see through the confusion. Good luck!
 
  • #5
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Ok thanks anyways. :smile:
 

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