Acceleration as a function of displacement

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!

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

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.
• 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? Feb 17 at 13:12