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.

Solution Steps

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:

$$v(t) = -32t + 5t^{-0.5}$$

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:

$$v(1.1) = -35.2 + 4.7675 = -30.4325$$

Final Answer