We are given the equation:

$t = \sqrt{\frac{d}{4.9}}$

where $t$ is the time in seconds, and $d$ is the distance traveled in meters by an object falling from rest. We need to determine the distance $d$ when the time $t = 12$ seconds.

First, rearrange the formula to solve for $d$:

$t^2 = \frac{d}{4.9}$

Multiply both sides by 4.9:

$d = 4.9 \times t^2$

Now substitute $t = 12$:

$d = 4.9 \times (12)^2 = 4.9 \times 144 = 705.6 \, \text{meters}$

So, the object travels 705.6 meters in 12 seconds.

Would you like more details on this solution, or have any questions? Here are 5 related questions you might explore:

1. How would the distance change if the time was halved?
2. What is the physical significance of the constant 4.9 in this equation?
3. How is the formula derived from the laws of motion?
4. What would the distance be for 10 seconds?
5. How does the formula relate to the acceleration due to gravity?

Tip: This formula comes from the kinematic equation for free fall under constant gravity $g = 9.8 \, \text{m/s}^2$, where $4.9$ is half of $g$.