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# x = position at time t

x= t^{2}- 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 $=1\ unit+1\ unit=2\ unit$

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