Answer
$30.3$ cm/s
Work Step by Step
First, the positions at t=2.00s and at t=3.00 s
are $x(2.00)$ and $x(3.00)$, respectively.
Let $x_{23}$ be the midpoint of $x(2.00)$ and $x(3.00):$
$x_{23}=\displaystyle \frac{x(2.00)+x(3.00)}{2}$
$=\displaystyle \frac{9.75+1.50(2.00)^{3}+9.75+1.50(3.00)^{3}}{2}=36.0$ cm
To find the t when the particle is at $x_{23}=36.0$ cm, we solve for t:
$36=9.75+1.50t^{3}$
$1.50t^{3}=36.0-9.75$
$t^{3}=\displaystyle \frac{36.0-9.75}{1.50}$
$t=(\displaystyle \frac{36.0-9.75}{1.50})^{1/3}=2.596$ s
So, the instantaneous speed at this time is
(speed = $|v(t)|$
We have already stated in previous parts that
$v(t)=4.50t^{2}$ (cm/s)
$|v(t)|=|4.5(2.596)^{2}|$ cm/s$=30.3$ cm/s
