The position of a particle along x-axis at time t is given by $x=1+t-t^2.$ the distance travelled by the particle on first 2 second is

The correct option is B 3 m

The correct option is D

Given,

Trajectory of motion of particle is $x=1+t-t^2.$ So, this is the parabolic equation.

So, differentiate x with respect to time

$\frac{dx}{dt}=0+1-2t$

$0=1-2t$

$\Rightarrow t=\frac{1}{2}=0.5$

At $t=0.$ , we have to check $\dfrac{d^2x}{dt^2}.

Now, again differentiate $\dfrac{dx}{dt} with respect to t

$\frac{d^{2}x}{d{t}^2}=-2$ Which is $<0$

Hence, at $t=0.5$, particle reaches maximum velocity and also chanhinf the nature of motion [ after $t=0.5$ , particle is retardating ]

So, total distance can be found by

$S=|x(t=0.5)-x(t=0)|+|x(t=2)-x(t=0.5)|$

$=|1+0.5-0.25-1|+|1+2-4-1-0.5+0.25|$

$=0.75+|-2-0.25|$

$=0.75+2.25$

$=3m.$