Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 62

The motion's equation of a particle is $s = 2t^3 - 7t^2 + 4t + 1$, where $s$ is in meters and $t$ is in seconds.

a.) Find the velocity and acceleration as functions of $t$.
Given: $s = 2t^3 - 7t^2 + 4t + 1$

Solve for first derivative and second derivative of the given equation to get the velocity and
acceleration as functions of $t$ respectively.


$
\begin{equation}
\begin{aligned}
V(t) &= 2t^3 - 7t^2 +4t + 1\\
\\
V'(t)&= 2 \frac{d}{dt} (t^3) - 7 \frac{d}{dt}(t^2) + 4 \frac{d}{dt}(t) + \frac{d}{dt}(1) && \text{Derive each term}\\
\\
V'(t)&= (2)(3t^2)-(7)(2t)+(4)(1)+0 && \text{Simplify the equation}\\
\end{aligned}
\end{equation}
$

The Velocity of a particle as function of $t$ is $V'(t) = 6t^2 - 14t + 4$


$
\begin{equation}
\begin{aligned}
a(t) &= 6t^2 - 14t + 4\\
\\
a'(t)&= 6 \frac{d}{dt}(t^2) -14 \frac{d}{dt} (t) + \frac{d}{dt} (4) && \text{Derive each term}\\
\\
a'(t)&= (6)(2t)-(14)(1)+0\\
\end{aligned}
\end{equation}
$


The Acceleration of a particle as function of $t$ is $a'(t) = 12t - 14$

b.) Find the acceleration after 1 second.

$
\begin{equation}
\begin{aligned}
& \text{Given: }\\
& \phantom{x} & t &= 1 sec\\
& \text{Equation in part(a)} \\
& \phantom{x} & a(t) &= 12t-14
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
a(t) &= 12t - 14 && \text{Substitute the given value of time}\\
\\
a(1) &= 12(1) - 14 && \text{Simplify the equation}
\end{aligned}
\end{equation}
$


The acceleration after 1 second is $\displaystyle a = -2 \frac{m}{s^2}$

c.) Graph the position, velocity and acceleration functions.