Question

suppose that the equation of motion for a particle is s = sin(3πt), where s is in meters and t in seconds.
a. find the velocity and acceleration as functions of t
velocity at time t: v =
acceleration at time t: a =
b. find the acceleration after 1 second.
acceleration after 1 second:
c. find the acceleration (in absolute value) at the instant when the velocity is 0 m/s.
acceleration:

Explanation:

Step1: Recall velocity - displacement relation

Velocity $v(t)$ is the derivative of displacement $s(t)$ with respect to time $t$. Given $s(t)=\sin(3\pi t)$, by the chain - rule $\frac{d}{dt}\sin(u)=\cos(u)\cdot\frac{du}{dt}$, where $u = 3\pi t$ and $\frac{du}{dt}=3\pi$. So $v(t)=\frac{ds}{dt}=3\pi\cos(3\pi t)$.

Step2: Recall acceleration - velocity relation

Acceleration $a(t)$ is the derivative of velocity $v(t)$ with respect to time $t$. Since $v(t)=3\pi\cos(3\pi t)$, using the chain - rule $\frac{d}{dt}\cos(u)=-\sin(u)\cdot\frac{du}{dt}$ with $u = 3\pi t$ and $\frac{du}{dt}=3\pi$, we get $a(t)=\frac{dv}{dt}=- 9\pi^{2}\sin(3\pi t)$.

Step3: Find acceleration at $t = 1$

Substitute $t = 1$ into the acceleration function $a(t)$. When $t = 1$, $a(1)=-9\pi^{2}\sin(3\pi)$. Since $\sin(3\pi)=0$, then $a(1)=0$.

Step4: Find when velocity is 0

Set $v(t)=3\pi\cos(3\pi t)=0$. Then $\cos(3\pi t)=0$. We know that $\cos\theta=0$ when $\theta=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}$. So $3\pi t=(2n + 1)\frac{\pi}{2}$, which gives $t=\frac{2n + 1}{6},n\in\mathbb{Z}$.

Step5: Find acceleration when velocity is 0

Substitute $t=\frac{2n + 1}{6}$ into the acceleration function $a(t)=-9\pi^{2}\sin(3\pi t)$. When $t=\frac{2n + 1}{6}$, $3\pi t=(2n + 1)\frac{\pi}{2}$, and $\sin((2n + 1)\frac{\pi}{2})=\pm1$. So $|a| = 9\pi^{2}$.

Answer:

a. Velocity at time $t$: $v = 3\pi\cos(3\pi t)$
Acceleration at time $t$: $a=-9\pi^{2}\sin(3\pi t)$
b. Acceleration after 1 second: $0$
c. Acceleration: $9\pi^{2}$