Questions: A businessman dropped a coin from the top floor of his office building and it fell according to the formula S(t)=-16 t^2+10 t^0.5, where t is the time in seconds and S(t) is the distance in feet from the top of the building. What was the instantaneous velocity at t=1.1 seconds? Round your answer to 2 decimal places.

A businessman dropped a coin from the top floor of his office building and it fell according to the formula S(t)=-16 t^2+10 t^0.5, where t is the time in seconds and S(t) is the distance in feet from the top of the building.

What was the instantaneous velocity at t=1.1 seconds? Round your answer to 2 decimal places.

Transcript text: A businessman dropped a coin from the top floor of his office building and it fell according to the formula $S(t)=-16 t^{2}+10 t^{0.5}$, where $t$ is the time in seconds and $S(t)$ is the distance in feet from the top of the building. What was the instantaneous velocity at $t=1.1$ seconds? Round your answer to 2 decimal places.

Solution

Solution Steps

Step 1: Differentiate the Position Function to Find Velocity

To find the instantaneous velocity, we need to differentiate the position function $ S(t) = -16t^2 + 10t^{0.5} $ with respect to time $ t $. The derivative of $ S(t) $ with respect to $ t $ gives us the velocity function $ v(t) $.

\[ v(t) = \frac{d}{dt}(-16t^2 + 10t^{0.5}) \]

Using the power rule for differentiation, we have:

\[ v(t) = -32t + 5t^{-0.5} \]

Step 2: Evaluate the Velocity Function at $ t = 1.1 $

Now, substitute $ t = 1.1 $ into the velocity function to find the instantaneous velocity at that time.

\[ v(1.1) = -32(1.1) + 5(1.1)^{-0.5} \]

Calculate each term:

\[ -32(1.1) = -35.2 \]

\[ 5(1.1)^{-0.5} = 5 \times \frac{1}{\sqrt{1.1}} \approx 5 \times 0.9535 = 4.7675 \]

Add the results:

\[ v(1.1) = -35.2 + 4.7675 = -30.4325 \]