# A particle is moving along $x$-axis whose position is varying with time according to x=-3t+$t^3$. The displacement from t=1 to t=3 is

This question is as simple as the just substituting the values (that’s what I think, doesn’t have to be right). So we get

For $$t=1$$ $$x=-3+1=-2$$

For $$t=3$$ $$x=-9+27=18$$

Adding both we get $$+16$$ . The answer given is $$20$$ . Why is this so?

As others have noted, the answer is $$20$$ because the computation involves $$18 - (-2) = 20$$. More generally, displacement is defined as an object's final position minus its initial position. Note that this doesn't have anything to do with absolute values, actually—you can have a negative displacement!
The initial position of the particle is $$-2$$ and the final position is $$1$$. So displacement is $$final \;position - initial\; position$$ $$=18-(-2)$$ $$=18+2$$ $$=20$$
It's simple, you have to add the absolute values .The negative sign in coordinate of $$x$$ indicates that its 2 units away from origin in opposite direction. you should draw a diagram to see it clearly.