Question

A particle moves in a straight line. Its position (in $m$) as function of time is given by
$x=({at}^2+b)$
What is the average velocity in time interval $t=3s$ to $t=5s$ in ${ms}^{-1}$. (Where $a$ and $b$ are constants and $a=1{ms}^{-2}$, $b=1m$)

Answer 1

$a=1m{s}^{-2}b=1m$

$x\left(t\right)=t^2+1$

$x\left(3\right)=9+1=10m$

$x\left(5\right)=25+1=26m$

$v_{avg}$ from t = 3s to t = 5s $=\frac{x\left(5\right)-x\left(3\right)}{2}$

$v_{avg}=\frac{26-10}{2}=8m/s$