The motion's equation of a particle is $s=t^3-3t$, where $s$ is in meters and $t$ is in seconds.
a.) Find the velocity and aceleration as function of $t$
Given: $s = t^3-3t$
Take the 1st derivative of the given equation to get the velocity and 2nd derivative to get the acceleration.
$
\begin{equation}
\begin{aligned}
V(t) &= t^3 - 3t\\
V'(t)&= \frac{d}{dt}(t^3) - 3 \frac{d}{dt}(t) && \text{(Derive each term)}\\
V'(t)&= 3t^2-3(1) && \text{(Simplify the equation)}
\end{aligned}
\end{equation}
$
The velocity of a particle as function of $t$ is $V'(t) = 3t^2-3$
$
\begin{equation}
\begin{aligned}
a(t) &= 3t^2 - 3\\
a'(t)&= 3\frac{d}{dt} (t^2) - \frac{d}{dt}(3)\\
a'(t)&= (3)(2t) -0\\
\end{aligned}
\end{equation}
$
The acceleration of a particle as function of $t$ is $a'(t) = 6t$
b.) Find the acceleration after $2s$
Given: $a(t) = 6t \qquad t = 2$ sec.
$
\begin{equation}
\begin{aligned}
a(t) &= 6t && \text{Use the formula of acceleration in part(a)}\\
a(2) &= 6(2) && \text{Substitute the given time}
\end{aligned}
\end{equation}
$
The acceleration after $2s$ is $\displaystyle a = 12 \frac{m}{s^2}$
c.) Find the acceleration when the velocity is 0.
$
\begin{equation}
\begin{aligned}
& \text{Given: }\\
& \phantom{x} & V(t) &= 0\\
& \text{Equation in part(a):}\\
& \phantom{x} & V(t) &= 3t^2 -3\\
& \phantom{x} & a(t) &= 6t
\end{aligned}
\end{equation}
$
$
\begin{equation}
\begin{aligned}
V(t) &= 3t^2 - 3 && \text{Substitute the given Velocity}\\
\\
0 &= 3t^2 - 3 && \text{Add 3 to each sides}\\
\\
3t^2 &= 3 && \text{Divide both sides by 3}\\
\\
\frac{3t^2}{3} &= \frac{3}{3} && \text{Take the square root of both lines}\\
\\
\sqrt{t^2} &= \sqrt{1} && \text{Simplify the equation}\\
\\
t &= 1 && \text{Time when Velocity is 0}\\
\\
a(t) &= 6t && \text{Substitute the computed time when Velocity is 0}\\
\\
a(1) &= 6(1) && \text{Simplify the equation}
\end{aligned}
\end{equation}
$
When the velocity is 0, the value of the acceleration is $\displaystyle a = 6 \frac{m}{s^2}$
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