Question

A particle moves according to the equation $x=2t^2-5t+6$ where, $x$ is in $\text{m}$. Find
$\left(i\right)$ average velocity in the first $3\text{s}$, and
$\left(ii\right)$ instantaneous velocity at $t=3\text{s}$.

• A. $1\text{m/s},7\text{m/s}$
• B. $4\text{m/s},3\text{m/s}$
• C. $2\text{m/s},5\text{m/s}$
• D. $3\text{m/s},7\text{m/s}$

Answer 1

The correct option is A $1\text{m/s},7\text{m/s}$

Given, $x=2t^2-5t+6$
So, for case $\left(i\right)$ we have
$x_{|@t=3\text{s}}=2\times (3{)}^2-5\times 3+6=9\text{m}$
$x_{|@t=0\text{s}}=2\times (0{)}^2-5\times 0+6=6\text{m}$
$\therefore$ Average velocity in the first $3\text{s}$ is
$v_{av}=\frac{x_{|@t=3\text{s}}-x_{|@t=0\text{s}}}{3-0}=\frac{9-6}{3-0}=1\text{m/s}$
For case $\left(ii\right)$ we have
instant velocity, $v=\frac{dx}{dt}=\frac{d(2t^2-5t+6)}{dt}=4t-5$
So, velocity at $t=3\text{s}$ is
$v=4\times 3-5=7\text{m/s}$