Answer
- Velocity $=0m/s$
- Speed $=0m/s$
- Acceleration $=-\sqrt2m/s^2$
- Jerk $=0m/s^3$
Work Step by Step
$$s=f(t)=\sin t+\cos t$$
a) Velocity $v$:
$$v(t)=f'(t)=(\sin t+\cos t)'=\cos t-\sin t$$
So, $$v(\pi/4)=\cos(\pi/4)-\sin(\pi/4)=\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=0(m/s)$$
b) Speed:
The speed at time $t=\pi/4$ $=|v(\pi/4)|=0(m/s)$.
c) Acceleration $a$:
$$a(t)=v'(t)=(\cos t-\sin t)'=-\sin t-\cos t$$
So, $$a(\pi/4)=-\sin(\pi/4)-\cos(\pi/4)=-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2(m/s^2)$$
d) Jerk $j$:
$$j(t)=a'(t)=(-\sin t-\cos t)'=-\cos t+\sin t$$
Therefore, $$j(\pi/4)=-\cos(\pi/4)+\sin(\pi/4)=-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=0(m/s^3)$$
