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I am given a question such that a 0.280kg object has a displacement (in meters) of $x=5t^3-8t^2-30t$. I need to find the average net power input from the interval of $t=2.0s$ to $t=4.0s$.

I know the formula for average net power is $\frac{\int^{x_2}_{x_1}F \ dx}{t_2-t_1}$ as the force (acceleration is not constant). The acceleration is given by $a = 30t-16$. The force is then given by $F=ma$, but since $m$ is a constant, I intend to ignore it in my calculations. As such, all I need to do is to express acceleration as a function of displacement $x$.

I initially tried to substitute $t=\frac{a+16}{30}$ into the displacement equation, but ended up with a complex expression in $a$ that I could not integrate $x$ against.

I then attempted to try chain rule, with $\frac{da}{dx} = \frac{da}{dt} \div\frac{dx}{dt} =\frac{30}{15t^2-16t-30}$, but this is an expression in $t$ and I still cannot perform $\int^{x_2}_{x_1} F \ dx$.

Does anyone have any advice on what I can do? Many thanks for any help extended!

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    $\begingroup$ Try this $Fdx=F\dfrac{dx}{dt}dt=mavdt$ $\endgroup$
    – Eli
    Feb 17 at 13:00

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You can use the fact that $\frac{dx}{dt} = 15t^{2} -16t -30$ first of all. Then, you can make a substitution into your work done integral, for $dt$, and change the limits so that instead of the displacement $x_{i}$, you have whatever initial and final times $t_{i}$. That should then work! If not, let me know and I can give you an explicit answer, but do try it yourself first.

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    $\begingroup$ Thank you so much. I kinda get what you mean, essentially I will replace $dx$ with $dx=15t^2-16t-30 \ dt$. In the replacement of the limits, does that mean I can simply put in $t=2$ and $t=4$? Is my understanding correct? $\endgroup$
    – a9302c
    Feb 17 at 13:12
  • $\begingroup$ Yep, exactly right! If you're happy with my response you can accept it as answer :) $\endgroup$
    – jambajuice
    Feb 17 at 14:11
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    $\begingroup$ Done! Thank you so much for the assistance! $\endgroup$
    – a9302c
    Feb 18 at 3:33

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