1. Joe bikes $x$ miles East at $20$ mph to his friend’s house. The time taken for this leg of the journey is:
$$t_1 = \frac{x}{20}$$

2. Joe then turns South and bikes $x$ miles at $20$ mph to the store. The time taken for this leg of the journey is:
$$t_2 = \frac{x}{20}$$

3. Joe then turns East again and bikes to his grandma’s house at $14$ mph. The distance from the store to his grandma’s house is $x + 2$ miles. The time taken for this leg of the journey is:
$$t_3 = \frac{x + 2}{14}$$

4. Joe spends a total of 1 hour on the bike to get to his grandma’s house. Therefore, we have:
$$t_1 + t_2 + t_3 = 1$$
Substituting the expressions for $t_1$, $t_2$, and $t_3$:
$$\frac{x}{20} + \frac{x}{20} + \frac{x + 2}{14} = 1$$
Simplifying the equation:
$$\frac{2x}{20} + \frac{x + 2}{14} = 1$$
$$\frac{x}{10} + \frac{x + 2}{14} = 1$$
To solve for $x$, find a common denominator (which is 70):
$$\frac{7x}{70} + \frac{5(x + 2)}{70} = 1$$
$$\frac{7x + 5x + 10}{70} = 1$$
$$\frac{12x + 10}{70} = 1$$
$$12x + 10 = 70$$
$$12x = 60$$
$$x = 5$$

5. The distance from Joe’s grandma’s house to his home is the hypotenuse of a right triangle with legs $5$ miles and $12$ miles (since $x = 5$ and $x + 2 = 7$):
$$\text{Distance} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ miles}$$

6. Joe rides straight home in his grandma’s helicopter at $78$ mph. The time taken to get home is:
$$t = \frac{13 \text{ miles}}{78 \text{ mph}}$$
Converting this time to minutes:
$$t = \frac{13}{78} \times 60 = \frac{13 \times 60}{78} = 10 \text{ minutes}$$

The final answer is $\boxed{10}$