A rock is dropped off the edge of a cliff and its distance from the top of the cliff after t seconds is s(t) = 16t^2. When will the rock strike the water, and make a table of avera... A rock is dropped off the edge of a cliff and its distance from the top of the cliff after t seconds is s(t) = 16t^2. When will the rock strike the water, and make a table of average velocities as it falls. Read more

Steps to Solve

1. Find the time when the rock strikes the water

To find when the rock strikes the water, set the distance equation equal to 400 ft: $$ s(t) = 16t^2 = 400 $$ Now, solve for $t$: $$ t^2 = \frac{400}{16} = 25 $$ Taking the square root, $$ t = 5 \text{ seconds} $$

2. Set up the average velocity formula

The average velocity over an interval $[a - h, a]$ can be calculated using the formula: $$ \text{Average Velocity} = \frac{s(a) - s(a - h)}{h} $$ where $h$ is the time interval.

3. Calculate the average velocities for given intervals

Now calculate the average velocities for different time intervals:

• For $h = 1$: $$ \text{Average Velocity} = \frac{s(5) - s(4)}{1} = \frac{400 - 256}{1} = 144 , \text{ft/sec} $$

• For $h = 0.5$: $$ \text{Average Velocity} = \frac{s(5) - s(4.5)}{0.5} = \frac{400 - 324}{0.5} = 152 , \text{ft/sec} $$

• For $h = 0.1$: $$ \text{Average Velocity} = \frac{s(5) - s(4.9)}{0.1} = \frac{400 - 384.16}{0.1} = 158.4 , \text{ft/sec} $$

• For $h = 0.01$: $$ \text{Average Velocity} = \frac{s(5) - s(4.99)}{0.01} = \frac{400 - 399.6004}{0.01} = 39.64 , \text{ft/sec} $$

• For $h = 0.001$: $$ \text{Average Velocity} = \frac{s(5) - s(4.999)}{0.001} = \frac{400 - 399.996016}{0.001} = 3.984 , \text{ft/sec} $$

4. Approximate the velocity

As $h$ approaches $0$, the average velocity approximates the instantaneous velocity at $t = 5$. The instantaneous velocity can be found by deriving the distance equation: $$ v(t) = s'(t) = 32t $$ At $t = 5$: $$ v(5) = 32 \times 5 = 160 , \text{ft/sec} $$