A particle of mass $40$ kg moves in a straight line such that the force (in newtons) acting on it at time $t$ (in seconds) is given by $$160t^4-320t^2-360$$ If at time $t=0$ its velocity $v$ is given by $v(0)=10$, and its position $x$ (in m) is given by $x(0)=14$, what is the position of the particle at time $t$?
I have that by Newton's second Law $$F=m\cdot a=40\cdot\frac{dv}{dt}=160t^4-320t^2-360\to\frac{dv}{dt}=4t^4-8t^2-9\\ v(t)=\frac{4}{5}t^5-\frac{8}{3}t^3-9t+v_0$$
$v(0)=v_0=10$.
What do I need to substitute in and where to get the final answer?
Now, you know the velocity, you can use:
$v(t)=\dfrac{dx}{dt}$
So you compute the particle's position at time t: $x(t)$.
$x(t)=\dfrac{2}{15}t^6−\dfrac{2}{3}t^4−\dfrac{9}{2}t^2+10t+C$, where C is a constante.
Boundary condition:
$\begin{align} x(0) &= C\\ &=14 \end{align}$
Thus: $x(t)=\dfrac{2}{15}t^6−\dfrac{2}{3}t^4−\dfrac{9}{2}t^2+10t+14$
Note also: $a(t)=\dfrac{d^2x}{dt^2}$ is the acceleration of the particule at time t.