A baby crawls $8$ feet towards the west and then $8$ feet towards the north. It then moves $14$ feet towards the east. How far and in which direction is the baby from the starting point?

Determine the distance and direction of the baby from the starting point.

In the figure,$\mathrm{AB}=8$ is the path baby crawls toward the west,

$\mathrm{BC}=8$ is the path baby crawls toward the north after turning.

$\mathrm{CE}=14$is the path baby crawls toward east finally.

So, the path covered is $\mathrm{A}\stackrel{\mathrm{west}}{\to }\mathrm{B},\mathrm{B}\stackrel{\mathrm{north}}{\to }\mathrm{C}\mathrm{and}\mathrm{C}\stackrel{\mathrm{East}}{\to \mathrm{E}}$ as shown below:

$$\begin{gathered} \mathrm{C}-----\mathrm{D}---->\overline{)\mathrm{E}} \\ \uparrow ⋮⋰ \\ \mathrm{I}⋮⋰ \\ \mathrm{I}⋮⋰ \\ \mathrm{B}<----\overline{)\mathrm{A}}⋰ \end{gathered}$$

Distance between the initial and final point is $\mathrm{AE}$ toward North east direction.

Let $\mathrm{D}$ be any point in $\mathrm{CE}$ such that $\mathrm{CD}=8\mathrm{ft}$.

Since with respect to the direction of paths, $\mathrm{AB}\parallel \mathrm{CD}\mathrm{and}\mathrm{AD}\parallel \mathrm{BC}$ and also $\mathrm{AB}=\mathrm{CD}=\mathrm{BC}=\mathrm{AD}=8\mathrm{ft}$.

Thus a quadrilateral $\mathrm{ABCD}$ is obtained which is square.

And also a right-angled triangle $∆\mathrm{ADE}$ formed.

Apply Pythagoras theorem in right angle to find the distance from point $A$ to point $E$:

$\begin{array}{rcl}{\mathrm{AE}}^2& =& {\mathrm{AD}}^2+{\mathrm{DE}}^2\\ & \Rightarrow & {\mathrm{AE}}^2={\mathrm{AD}}^2+{\left(\mathrm{CE}-\mathrm{CD}\right)}^2\\ & \Rightarrow & {\mathrm{AE}}^2=8^2+{\left(14-8\right)}^2\\ & & \Rightarrow {\mathrm{AE}}^2=64+36\\ & \Rightarrow & {\mathrm{AE}}^2=100\\ & \Rightarrow & \mathrm{AE}=\sqrt{100}\\ & \Rightarrow & \mathrm{AE}=10\end{array}$

Therefore, distance between the starting point and ending point is $10\mathrm{m}$ in the north east direction.