Find the magnitude of the infant’s displacement.

$\Delta s=6.7\ \mathrm{m}$

Although the infant crawled a total distance of $5+3+1=9\ \mathrm{m}$, this is not the displacement, which is merely the net distance traveled.

Using the Pythagorean Theorem, we can calculate that the magnitude of the displacement is

$$\Delta s=\sqrt{\left(\Delta x\right)^2+\left(\Delta (y)\right)^2}$$

$$\Delta s=\sqrt{\left(6\ \mathrm{m}\right)^2+\left(3\ \mathrm{m}\right)^2}$$

$$\Delta s=\sqrt{45\ \mathrm{m^2}}$$

$$\Delta s=6.7\ \mathrm{m}$$