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A particle of mass m is initially at rest

  • Thread starter Calpalned
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  • #1
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Homework Statement


A particle of mass m, initially at rest at x=0, is accelerated by a force that increases in time as F=Ct2. Determine its velocity v as a function of time.

Homework Equations


x = vt
v = at

The Attempt at a Solution


The correct method makes sense, but my method has no error. However, my answer is wrong. Why?
http://photo1.ask.fm/726/204/412/-69996997-1sha200-dqjla0edmhpamrf/original/IMG_4066.jpg [Broken]
 
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Answers and Replies

  • #2
haruspex
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v=at only works for constant acceleration. Integration works for constant or varying acceleration.
 
  • #3
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So because force is a function of time (f = Ct2), acceleration is also a function of time and therefore not constant?
Thanks
 
  • #4
haruspex
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So because force is a function of time (f = Ct2), acceleration is also a function of time and therefore not constant?
Thanks
Yes.
 
  • #5
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Thank you so much
 
  • #6
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I don't see how you got 3m at the bottom? does it have to do with the fact that t is to the third power?
 
  • #7
haruspex
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I don't see how you got 3m at the bottom? does it have to do with the fact that t is to the third power?
Yes. What is ##\int t^2.dt##?
 
  • #8
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Yes. What is ##\int t^2.dt##?
I figured it out, it is the antiderivative I believe
 
  • #9
I solved this questions like;

$$ F=m.a $$ $$ F=Ct^2 $$ => $$ m.a = Ct^2 $$ $$ a= \frac {Ct^2} {m} $$
and if we integrate "a" respect to time(t), we obtain velocity(v);
$$ \int \frac {Ct^2} {m} dt = \frac {Ct^3} {3m} = V $$
and if we integrate "v" respect to time(t), we obtain position(x);
$$ \int \frac {Ct^3} {3m} dt = \frac {Ct^4} {12m} = X $$

I'm not sure, tell me if it's correct.
 
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  • #10
haruspex
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tell me if it's correct
It is, but technically you should go through the step of allowing for a constant of inrtegration each time, then using the initial conditions to show it is zero.
 

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