Particle acceleration and velocity Problem

  • Thread starter s4orce
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  • #1
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A particle is moving along a straight line such that its acceleration is defined as a=(4s^2)m/s^2, where s is in meters. If v=-100m/s when s=10m and t=0, determine the particles velocity as a function of position.

Now I'm taking the integral of a but when I plug in the other parts, it doesn't work out. I think I'm doing something wrong.

Someone please comment on how-to approach this problem step by step thanks!
 

Answers and Replies

  • #2
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[tex]a(t)= \frac{d(s')}{dt} =4s^2[/tex]

Move over dt integrate, set the limits, and solve.
 
  • #3
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Can you explain a little further please?
 
  • #4
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Which part?

What dynamics book are you using? There should be a problem similar to this given somewhere as an example.
 
  • #5
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after moving over dt to integrate, how did you approach it from there
 
  • #6
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what do you mean? Show me some work please.
 
  • #7
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how can i post the work here?
 
  • #8
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Type out what you got and what you did.

You can use [tex] and [ / tex] tags to make it look nicer.
 
  • #9
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integral 4s^2 = (4s^3)/3

s=10

4(10^3)/3 = 1333.33

1333.33/10 = 133.33 = particle velocity at x position
 
  • #10
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Are you familiar with the definition of acceleration?

Which class is this HW for?
 
  • #11
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It is for a dynamics class, I just want to get a head start on a set of problems that were given to me
 
  • #12
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I think you need to review the basics of dynamics. What it means when they say acceleration, velocity, and position.

I think you need to solidify the fundamentals before doing this HW. You should know better that acceleration is the second derivative of position by now.
 
  • #13
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so you cant help me?
 
  • #14
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Sure, tell me what the definition of acceleration is, and look for your mistake.

I can point out mistakes, but im not going to teach you fundamentals. :wink:
 
  • #15
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can you solve it and I can find my mistake, or post how.

also acceleration is the rate at which a object increases/changes its velocity(speed)
 
  • #16
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Ah, Im sorry. Its asking for velocity as a function of position. I was thinking you had to integrate twice.

What you need to do: set up your limits of integration.
 
  • #17
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?????????????????
 
  • #18
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Do you know what limits of integration are?
 
  • #19
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yes integral from one value to another
 
  • #20
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so what are your value you used?
 
  • #21
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0 to 10.............
 
  • #22
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What happened to the limits on the integral of the velocity?
 
  • #23
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-100x? that one?
 
  • #24
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When integrating, you always have two limits. Two for position, two for velocity. What are the two limits for the integral of velocity?
 
  • #25
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im confused man...
 
  • #26
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[tex] \int^v_{-100}[/tex]

Remember, I said you must have an upper and lower limit of integration.
 
  • #27
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can you just show me how you approac it please?
 
  • #28
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Sure, give me a second. I made another mistake reading the problem. Sorry!
 
  • #29
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ok thanks!
 
  • #30
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Ok, it was tricky. Had to crack open my dynamics book from a few semesters ago!

Here is the trick:

[tex] V=\frac{ds}{dt}[/tex]

Also,

[tex] a=\frac{dv}{dt}[/tex]

If you solve each of these for [tex]\frac{1}{dt}[/tex] you will get:

a*ds=v*dv

Now set your limits of integration and replace your relation for acceleration -a

[tex] \int^s_{10} 4s^2 ds = \int^V_{-100}vdv [/tex]

Is this clear?
 
Last edited:

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