The position of a particle along x-axis at time $t$ is given by $x=1+t-t^2\text{m}$. The distance travelled by the particle in first $2\text{sec}$ is

The correct option is C $2.5\text{m}$
$x=1+t-t^2$
At $t=0\text{sec},x_0=1\text{m}$
We know that, $v=\frac{dx}{dt}=1-2t$
So, at $t=\frac{1}{2}\text{sec},v=0$
And displacement , $x_{\frac{1}{2}}=1+\frac{1}{2}-\frac{1}{4}=\frac{5}{4}=1.25\text{m}$
Distance travelled by the particle from $t=0$ to $t=\frac{1}{2}\text{sec}$
$=|x_{\frac{1}{2}}-x_0|=1.25\text{m}-1\text{m}=0.25\text{m}$
Position of particle at the end of $2\text{sec}$ is given by
$x_2=1+2-2^2=-1\text{m}$
Distance travelled by the particle $t=\frac{1}{2}\text{sec}$ to $t=2$ sec
$=|x_2-x_{\frac{1}{2}}|=|-1m-1.25\text{m}|=2.25\text{m}$
Total distance travelled by the particle from $t=0$ sec to $t=2$ sec will be
$\therefore x=0.25\text{m}+2.25\text{m}=2.5\text{m}$