Answer
a. $5\sqrt 2$
b. $-10$ (furthest to the left), $10$ (furthest to the right).
c. $v(\frac{3\pi}{4})=0, a(\frac{3\pi}{4})=10$ and $v(\frac{7\pi}{4})=0, a(\frac{7\pi}{4})=-10$.
d. $t=\frac{\pi}{4}$, $v(\frac{\pi}{4})=-10$, $|v(\frac{\pi}{4})|=10$, $a(\frac{\pi}{4})=0$.
Work Step by Step
a. Given the equation of the particle $s(t)=10cos(t+\frac{\pi}{4}), t\leq0$, the starting position can be found as $s(0)=10cos(0+\frac{\pi}{4})=5\sqrt 2$
b. To find the extrema in $s(t)$, take the derivative to get $v(t)=s'(t)=-10sin(t+\frac{\pi}{4})$. Let $s'(t)=0$; we get $t=k\pi-\frac{\pi}{4}$ where $k$ is an integer and $k\geq1$. Use $k=1$ to get the minimum $s(\frac{3\pi}{4})=-10$ (furthest to the left). Use $k=2$ to find the maximum $s(\frac{7\pi}{4})=10$ (furthest to the right).
c. Taking the second derivative, we can get $a(t)=v'(t)=-10cos(t+\frac{\pi}{4})$. Using the time from part (b), we get the velocity and acceleration as $v(\frac{3\pi}{4})=0, a(\frac{3\pi}{4})=10$ and $v(\frac{7\pi}{4})=0, a(\frac{7\pi}{4})=-10$.
d. The minimum of $t$ to give $s(t)=0$ can be found as $t=\frac{\pi}{4}$ and we get velocity $v(\frac{\pi}{4})=-10$, speed $|v(\frac{\pi}{4})|=10$, and acceleration $a(\frac{\pi}{4})=0$.
