A stone fell from the top of a cliff into the ocean. $\newline$ In the air, it had an average speed of $16 \mathrm{~m} / \mathrm{s}$ . In the water, it had an average speed of $3 \mathrm{~m} / \mathrm{s}$ before hitting the seabed. The total distance from the top of the cliff to the seabed is $127$ meters, and the stone's entire fall took $12$ seconds. $\newline$ How long did the stone fall in the air and how long did it fall in the water?

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Solve equations with variables on both sides: word problems

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Q. A stone fell from the top of a cliff into the ocean.

\newline

In the air, it had an average speed of

16 \mathrm{~m} / \mathrm{s}

. In the water, it had an average speed of

3 \mathrm{~m} / \mathrm{s}

before hitting the seabed. The total distance from the top of the cliff to the seabed is

127

meters, and the stone's entire fall took

12

seconds.

\newline

How long did the stone fall in the air and how long did it fall in the water?

Set up equations: Set up the equations based on the given information. $\newline$ Let $t_{\text{air}}$ be the time the stone falls in the air and $t_{\text{water}}$ be the time it falls in the water. We know that the total time for the fall is $12$ seconds, so we can write the following equation: $\newline$ $t_{\text{air}} + t_{\text{water}} = 12$
Express total distance: Express the total distance fallen in terms of the distances in air and water. $\newline$ Let $d_{\text{air}}$ be the distance the stone falls in the air and $d_{\text{water}}$ be the distance it falls in the water. The total distance is given as $127$ meters, so we can write the following equation: $\newline$ $d_{\text{air}} + d_{\text{water}} = 127$
Use average speeds: Use the average speeds to relate time and distance for each part of the fall. $\newline$ The average speed in the air is $16\,\text{m/s}$ , and the average speed in the water is $3\,\text{m/s}$ . We can write two equations based on the definition of average speed (distance = speed $\times$ time): $\newline$ $d_{\text{air}} = 16 \times t_{\text{air}}$ $\newline$ $d_{\text{water}} = 3 \times t_{\text{water}}$
Substitute expressions: Substitute the expressions for $d_{\text{air}}$ and $d_{\text{water}}$ into the total distance equation. $\newline$ Using the equations from Step $3$ , we substitute the expressions for $d_{\text{air}}$ and $d_{\text{water}}$ into the total distance equation: $\newline$ $16 \cdot t_{\text{air}} + 3 \cdot t_{\text{water}} = 127$
Solve system of equations: Solve the system of equations to find $t_{\text{air}}$ and $t_{\text{water}}$ . We now have a system of two equations with two unknowns: $1$ ) $t_{\text{air}} + t_{\text{water}} = 12$ $2$ ) $16 \cdot t_{\text{air}} + 3 \cdot t_{\text{water}} = 127$ We can solve this system using substitution or elimination. Let's use substitution by expressing $t_{\text{water}}$ in terms of $t_{\text{air}}$ from the first equation: $t_{\text{water}} = 12 - t_{\text{air}}$
Find $t_{\text{air}}$ : Substitute $t_{\text{water}}$ into the second equation and solve for $t_{\text{air}}$ . $\newline$ Substituting $t_{\text{water}}$ into the second equation gives us: $\newline$ $16 \cdot t_{\text{air}} + 3 \cdot (12 - t_{\text{air}}) = 127$ $\newline$ Expanding this, we get: $\newline$ $16 \cdot t_{\text{air}} + 36 - 3 \cdot t_{\text{air}} = 127$ $\newline$ Combining like terms, we have: $\newline$ $13 \cdot t_{\text{air}} = 127 - 36$ $\newline$ $13 \cdot t_{\text{air}} = 91$ $\newline$ Dividing both sides by $13$ gives us: $\newline$ $t_{\text{air}} = \frac{91}{13}$ $\newline$ $t_{\text{water}}$ $0$
Find $t_{\text{water}}$ : Use the value of $t_{\text{air}}$ to find $t_{\text{water}}$ . Now that we have $t_{\text{air}}$ , we can find $t_{\text{water}}$ using the equation from Step $5$ : $t_{\text{water}} = 12 - t_{\text{air}}$ $t_{\text{water}} = 12 - 7$ $t_{\text{water}} = 5$