These problems are so much easier to write out if you do some trig to begin with. Here the 3 and the 4 are a bit of a giveaway indicating a 3,4,5 triangle. Taking the liberty of changing the d to the more usual s,
$\displaystyle s=3\sin(40\pi t)-4\cos(40\pi t)=5\left(\frac{3}{5}\sin(40\pi t)-\frac{4}{5}\cos(40\pi t)\right)$
$\displaystyle =5\sin(40\pi t-\alpha),$ where $\displaystyle \cos\alpha=3/5, \sin\alpha=4/5, \tan\alpha=4/3.$
That doesn't get you pass the calculus however.
Velocity$\displaystyle v=\frac{ds}{dt}=200\pi\cos(40\pi t-\alpha),$
Acceleration $\displaystyle a=\frac{dv}{dt}=-8000\pi^{2}\sin(40\pi t-\alpha).$
The particle will be at rest when the velocity is zero, that is, when $\displaystyle 200\pi\cos(40\pi t-\alpha)=0,$ etc.
seems you are in way over your head ... recommend you visit the links
Trigonometry | Khan Academy Video Course
Calculus | Khan Academy Video Course
In the end you will just understand this question, but when thrown another question similar to this you will get it wrong..It's ok I think I may be close to understanding.
Trust me I am talking from personal experience, it never pays.
You should watch those Khan videos, the guy is brilliant.
Here is another link to the physics Section:
Khan Academy