- #1

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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!

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- Thread starter s4orce
- Start date

- #1

- 39

- 0

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!

- #2

- 3,042

- 16

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

Move over dt integrate, set the limits, and solve.

Move over dt integrate, set the limits, and solve.

- #3

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Can you explain a little further please?

- #4

- 3,042

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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.

You can use [tex] and [ / tex] tags to make it look nicer.

- #9

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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?

Which class is this HW for?

- #11

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- #12

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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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I can point out mistakes, but im not going to teach you fundamentals.

- #15

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also acceleration is the rate at which a object increases/changes its velocity(speed)

- #16

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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

- 3,042

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so what are your value you used?

- #21

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0 to 10.............

- #22

- 3,042

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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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- #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.

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

- 3,042

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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?

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?

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