Deriving acceleration from Position vector help

  • Thread starter kpadgett
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Homework Statement


An object has a position given by r-> = [2.0 m + (1.00 m/s)t] i + [3.0m−(5.00 m/s2)t2] j, where all quantities are in SI units. What is the magnitude of the acceleration of the object at time t = 2.00 s?


Homework Equations


All I am thinking here is that I can find acceleration by deriving velocity from the position vector, then taking the deriving the velocity vector.


The Attempt at a Solution


deriving the original r vector for velocity gave me V=[2.0m + (1.00 m/s)(1)]i + [3.0m - (5.00 m/s^2)(2t)]j

deriving the velocity vector for a gave me A=[2.0m + (1.00 m/s)]i + [3.0m - (5.00 m/s^2)(t)]j

once I have this vector for acceleration, I get the acceleration at t=2.00s to be A(2.00s)=(3m)i + (-7m/s^2)j.

I find the vector A=7.616 m/s^2, which my online software says is incorrect. Can I get some input here?
 

Answers and Replies

  • #2
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What is the derivative of 2.0?
 
  • #3
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What is the derivative of 2.0?
Isn't the derivative of 2, 1? I apologize, I thought you left variables in the equation until you were done deriving? As in I would put 2 in for t after I had the equation for acceleration?
 
  • #4
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Your function is basically ## \vec{r}(t) = (a + bt)\vec{i} + (c + dt^2)\vec{j}##, where ## a, \ b, \ c, \ d ## are some constants. You differentiate the vector function component-wise, so you compute ## \frac {d} {dt} (a + bt) ## and ## \frac {d} {dt} (c + dt^2) ##.

Now, ## \frac {d} {dt} (a + bt) = \frac {d} {dt} a + \frac {d} {dt} bt##, and what is ## \frac {d} {dt} a ## if ## a ## is constant?
 
  • #5
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Wouldn't the derivative of a be 0?

So then I can find that V= [(0m)+(t)]i + [(0m)-(t^2)]j

So V= (t)i + (-t^2)

Deriving V, I get that A= (1)i + (-2t)j ? I see my answer here is wrong as well.... I really how no clue what I am doing wrong, and I can't even find deriving V and A from an R vector.
 
  • #6
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Ok, I now see here that the derivative of dt^2= 2dt. I'm going to see if this helps me out.
 
  • #7
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Wouldn't the derivative of a be 0?
Correct.

So then I can find that V= [(0m)+(t)]i + [(0m)-(t^2)]j
You have fixed the problem with constants. Now fix the problem with non-constants. What are ## \frac {d} {dt} (pt) ## and ##\frac {d} {dt} (qt^2) ##, where ## p, \ q ## are constant?
 
  • #8
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Correct.



You have fixed the problem with constants. Now fix the problem with non-constants. What are ## \frac {d} {dt} (pt) ## and ##\frac {d} {dt} (qt^2) ##, where ## p, \ q ## are constant?
d/dt (pt) woud be (t)? correct? and d/dt (q(t^2)) would be 2qt? so my vector for velocity is V= (t)i + [(-10m/s^2)t]j?

Giving the vector A= [d/dt (t)]i + [d/dt (-10(t)m/s^2)]

what I can see:
1)The derivative of t would be 1, as with any lone variable?
2)the derivative of 10(t) is t, since we just remove the constant>

so A= (1m)i - (tm/s^2)
 
  • #9
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d/dt (pt) woud be (t)? correct?
2)the derivative of 10(t) is t, since we just remove the constant
This is one error that plagues your calculations repeatedly. Review this subject.
 
  • #10
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This is one error that plagues your calculations repeatedly. Review this subject.
Wow, it was so simple that I never even thought about it. I don't know how id made it through math 137 and still make that mistake. Thank you very much.
 

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