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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 \( g \).

The formula \( t = \sqrt{\frac{d}{4.9}} \) gives the approximate distance \( d \) in meters traveled in \( t \) seconds by an object falling from rest. What is the distance an object can travel from rest in 12 sec?

Learn how to calculate the distance an object can travel from rest in 12 seconds using the formula \( t = \sqrt{\frac{d}{4.9}} \). This problem explores algebraic and kinematic concepts, making it suitable for students in Grades 9-10.

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