Question

Position of a particle in $m$ with moving along x-axis is given by $x=2+8t-4t^2$ where $t$ in $s$. The distance traveled by the particle from $t=0tot=2s$ is

• A. $0\text{m}$
• B. $8\text{m}$
• C. $12\text{m}$
• D. $16\text{m}$

Answer 1

The correct option is B $8\text{m}$

$x=2+8t-4t^2$
so $v=\frac{dx}{dt}=8-8t$
$v=0thent=1s$
After $t=1$, direction of velocity changes.
path is $x_{t=0}to{x}_{t=1s}$
then $x_{t=1s}to{x}_{t=2s}$
here $x_{t=0}=2m{x}_{t=1s}=6m,x_{t=2s}=2\text{m}$
Distance from $2\text{m}$ to $6\text{m}=\text{4}\text{m}$
Distance from $6\text{m}$ to $2\text{m}=4\text{m}$
$\text{Total distance}=8\text{m}$