When will the rock strike the water? Let t = a be the time when the rock strikes the water. Make a table of average velocities and approximate the velocity at which the rock strike... When will the rock strike the water? Let t = a be the time when the rock strikes the water. Make a table of average velocities and approximate the velocity at which the rock strikes the water.
Understand the Problem
The question is asking when a rock, dropped from a height of 400 feet, will hit the water based on the equation of motion s(t) = 16t^2. Part a requires a specific time when it will strike the water, and part b requires creating a table to calculate average velocities as time intervals approach that moment.
Answer
The rock will strike the water at $5$ seconds.
Answer for screen readers
The rock will strike the water at $5$ seconds.
Steps to Solve
- Set the distance equation equal to 400 ft
The distance from the top of the cliff to the water is given as 400 ft. We set the distance equation equal to this value:
$$ 16t^2 = 400 $$
- Solve for time, t
To find the time when the rock strikes the water, divide both sides of the equation by 16:
$$ t^2 = \frac{400}{16} $$
$$ t^2 = 25 $$
Now, take the square root:
$$ t = \sqrt{25} = 5 $$
- Result for part a
The rock will strike the water at $ t = 5 $ seconds.
- Define the average velocity formula
Average velocity is calculated using the formula:
$$ v_{avg} = \frac{s(b) - s(a)}{b - a} $$
where $s(t)$ is the position function.
- Calculate average velocities for various intervals
Use $a = 5$ and calculate average velocities for the following time intervals:
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Interval [4, 5] $$ v_{avg} = \frac{s(5) - s(4)}{5 - 4} = \frac{400 - 256}{1} = 144 \text{ ft/sec} $$
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Interval [4.5, 5] $$ v_{avg} = \frac{s(5) - s(4.5)}{5 - 4.5} = \frac{400 - 324}{0.5} = 152 \text{ ft/sec} $$
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Interval [4.99, 5] $$ v_{avg} = \frac{s(5) - s(4.99)}{5 - 4.99} = \frac{400 - 399.6001}{0.01} = 39.9 \text{ ft/sec} $$
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Interval [4.999, 5] $$ v_{avg} = \frac{s(5) - s(4.999)}{5 - 4.999} = \frac{400 - 399.960001}{0.001} = 39.999 \text{ ft/sec} $$
The rock will strike the water at $5$ seconds.
More Information
At $t = 5$ seconds, the rock reaches the water after falling from the cliff, and the calculated average velocities show that as the time intervals get smaller, the average velocity approaches a specific value, indicating the instantaneous velocity at that point.
Tips
- Not squaring $t$ when solving the equation. Make sure to handle square roots correctly.
- Confusing average velocity with instantaneous velocity; remember that the average velocity is over intervals, while instantaneous is at a single point.
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