Given velocity of a particle v = t-t2 ............[1]
If x represents displacement then v=(dx /dt)= (t-t2)
Integrating we get x = t2/2-t3/3+c , c is the integration constant
Imposing the condition that at t=0 ,x=0 we get c= 0 hence we have x = t2/2-t3/3.........[2]
So putting t=2 we get net displacement in 2 sec will be [x]t=2= 22/2-23/3= -2/3 unit
Now putting v=0 in the given equation we get, 0 = t-t2 => t= 0 and t-1=0
So initially the velocity is zero. and it is also zero after 1sec.
Putting t=1 in equation [2] we get distance traveled during this 1 sec as x1 = 1/2-1/3= 1/6 unit
after 1 sec particle begins to retrace its path. Now putting x=0 in [2] we get t =0 and t=3/2
So it reaches at origin in next (3/2-1)=1/2 sec traveling distance x2 = 1/6 unit
and finally in last 1/2 sec it travels x3= 2/3 unit.
So we can say total distance traveled in 2 sec is =x1+x2+x3=1/6+1/6+2/3= 1 unit