Question

The position of an elevator varies with time according to the function $x\left(t\right)=t^2+2t+3$

Find the
$\left(1\right)$ average velocity between time interval $t=0\text{s}$ to $t=2\text{s}$.
$\left(2\right)$ instantaneous velocity of the elevator at time $t=3\text{s}$.

• A. $v_{avg}=8\text{m/s}$ ; $v_{ins}=11\text{m/s}$
• B. $v_{avg}=4\text{m/s}$ ; $v_{ins}=8\text{m/s}$
• C. $v_{avg}=8\text{m/s}$ ; $v_{ins}=4\text{m/s}$
• D. $v_{avg}=11\text{m/s}$ ; $v_{ins}=4\text{m/s}$

Answer 1

The correct option is B $v_{avg}=4\text{m/s}$ ; $v_{ins}=8\text{m/s}$

Given: $x\left(t\right)=t^2+2t+3$
So, at $t=0\text{s}$ ; $x_0=3\text{m}$
and at $t=2\text{s}$ ; $x_2=4+4+3=11\text{m}$
$\therefore v_{avg}=\frac{x_3-x_0}{2}=\frac{11-3}{2}=4\text{m/s}$

Now,
$v_{ins}=Slope=\frac{dx}{dt}=\frac{d}{dt}(t^2+2t+3)$
$\Rightarrow v_{ins}=2t+2$
$\therefore v_{ins}(t=3\text{s})=2\times 3+2=8\text{m/s}$