Answer
a. $10cm/sec$, $t=\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}$ sec. $0cm$, $0cm/s^2$
b. $t=0,1,2,3,4$ sec, $10cm$ $0cm/sec$
Work Step by Step
a. Given the equation for the cart position $s(t)=10cos\pi t, 0\leq t\leq 4$, we have $v(t)=s'=-10\pi\ sin\pi t$ and $a(t)=s''=-10\pi^2 cos\pi t$. The maximum speed can be found as $|v(\frac{\pi}{2})|=10cm/sec$. This maximum happens when $s''=0$, which gives $t=\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}$ sec. The position at these times is $s(t)=10cos\pi t=0$ and the magnitude of the acceleration at these times is $|a(t)|=|-10\pi^2 cos\pi t|=0cm/s^2$
b. The magnitude of the acceleration is the greatest when the jerk $j(t)=0$. Since $j(t)=a'(t)=10\pi^3sin\pi t=0$, we have $t=0,1,2,3,4$ sec and $|a(0)|=10\pi^2 cm/s^2$. The position of the cart at these times is $s(t)=\pm10cm$ and the amplitude is $|S(t)|=10cm$, which is $10cm$ from its rest position. The speed at these times is $v(t)=-10\pi\ sin\pi t=0$
