The correct option is C 2.5 m
x=1+t−t2
At t=0 sec, x0=1 m
We know that, v=dxdt=1−2t
So, at t=12 sec,v=0
And displacement, x1/2=1+12−14=54=1.25 m
Distance travelled by the particle from t=0 to t=12 sec
=∣∣x1/2−x0∣∣=1.25 m−1 m=0.25 m
Position of particle at the end of 2 sec is given by
x2=1+2−22=−1 m
Distance travelled by the particle t=12 sec to t=2 sec
=∣∣∣x2−x12∣∣∣=|−1 m−1.25 m|=2.25 m
Total distance travelled by the particle from t=0 sec to t=2 sec will be
∴x=0.25 m+2.25 m=2.5 m