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x = position at time t
x= t2- 2t
Find distance and displacement in first 1 sec and 2 sec.
As given, $x=$position at $t$ time
$x=t^2-2t$
So, velocity $v=\frac{dx}{dt}=2t-2$
Let's check the sign of velocity at different time intervals.
At $t=0$, $v=-2\ unit/s$
At $t=1\ s$, $v=2\times1-2=0\ unit/s$
At $t=2\ s$, $v=2\times2-2=2\ unit/s$
At $t=0$, $x=0$
At $t=1\ s$, $x=1^2-2\times 1=1-2=-1$
At $t=2\ s$, $x=2^2-2\times2=4-4=0$
Thus, we can say that at first $1\ s$, the object moves $1\ unit$ towards the negative x-axis. And moves back towards the positive x-axis at first $2\ s$ as shown in the graph.
So, displacement at first $1\ s=$change in position $=1\ unit$
Distance traveled by the object at first $1\ s=1\ unit$
At first $2\ s$, $x=0$ it indicates that the object moves back to $0$ towards the positive x-axis.
So, displacement $=0$, as the object return to its initial position.
at first $2\ s$, distance traveled $=$total length of the path traveled $=1\ unit+1\ unit=2\ unit$