- #1

mikejones2000

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I found the acceleration by double derivation of the function and then multiplied the function by 3 to give me the force and then integrated but that did not work. Any help is greatly appreciated.

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- Thread starter mikejones2000
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- #1

mikejones2000

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I found the acceleration by double derivation of the function and then multiplied the function by 3 to give me the force and then integrated but that did not work. Any help is greatly appreciated.

- #2

Hootenanny

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You didn't need to integrate, just multiply the force you found by the distance. And allow me to take a guess that you integrated with respect to t? This is incorrect as the defintion of work is the integral of force with respect to displacement.

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- #3

mikejones2000

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- #4

Hootenanny

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[tex]P(t) = F(t)\cdot V(t)[/tex]

Now,

[tex]\text{Work} = \int P(t) \; dt[/tex]

Can you go from here?

- #5

StatusX

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- #6

mikejones2000

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- #7

Hootenanny

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[tex]\text{Work} = \int^{3}_{0} |F(t)|\cdot V(t)[/tex]

Since, as Status points out, although work is being done in the negative direction, work is still being done by the force, work is a scalar not a vector. Technically you are taking two vectors, (force and displacement) and turning them into a scalar by using the dot product.

- #8

mikejones2000

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- #9

arunbg

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In this problem, we consider forces and velocities to be along only one direction and there is no need for dot products or even vectors for that matter .

Remember velocity, V = dx/dt .

Reread all the previous posts. The approach has been outlined clear enough .

- #10

Hootenanny

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[tex]\text{work} = \int^{3}_{0} |F(t)| \cdot v(t) \; dt[/tex]

Which can be written as;

[tex]\text{work} = \int^{3}_{0} \left| m\cdot\frac{d^{2}x}{dt^2} \right| \cdot\frac{dx}{dt} \; dt[/tex]

Does that make it more obvious?

- #11

mikejones2000

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x=3t^3-4t^2+3t (dx/dt)=3t^2-8t+3 (|d^2x/d^2t|)=6t+8.

3(6t+8)=18t+24=F V=3t^2-8t+3

(18t+24)(3t^2-8t+3)=54t^3-120t^2-138t+72=P(t)

Integration of P(t) yields=(54t^4/4)-(120t^2/3)-(138t^2)/(2)+72t) from 0 to 3 seconds.

- #12

StatusX

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mikejones2000 said:(dx/dt)=3t^2-8t+3 (|d^2x/d^2t|)=6t+8.

This step isn't right. To take the absolute value, you'll have to find where the function becomes negative, and split it up around this point. For example:

[tex] |x-3|= \left\{ \begin{array}{cc} x-3 & \mbox{ if } x>3 \\ -(x-3) & \mbox{ otherwise} \end{array}[/tex]

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- #13

mikejones2000

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{6t-8 if x>4/3

so |6t-8|={-(6t-8) otherwise.

I am not really sure how to run with this

so |6t-8|={-(6t-8) otherwise.

I am not really sure how to run with this

- #14

StatusX

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- #15

mikejones2000

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- #16

Hootenanny

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No, you would do the two whole integrals seperatly, thus;mikejones2000 said:

[tex]\int^{\frac{4}{3}}_{0} [ m(8-6t) ] \cdot (3t^{2} - 8t + 3) \; dt[/tex]

[tex]\int^{3}_{\frac{4}{3}} [ m(6t - 8) ] \cdot (3t^{2} - 8t + 3) \; dt[/tex]

- #17

petergel

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ie:

acceleration is

I take the 3 out of the integral to make things simpler.

Also we now have limits of integration of 0.0s to 3.0sec

This gives us...

// Now we substitute 0.0sec to 3.0 sec for t and should get our answer.

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