Exponential and Logarithmic Functions
Finding the time given an exponential function with base e that models a...
Antonio
Suppose that the velocity $v(t)$ (in meters per second) of a sky diver falling near the Earth's surface is given by the following exponential function, where time $t$ is the time after diving measured in seconds.
\[
v(t)=55-55 e^{-0.24 t}
\]

How many seconds after diving will the sky diver's velocity be 41 meters per second?
Round your answer to the nearest tenth, and do not round any intermediate computations.
seconds
Explanation
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Step 1 ：Suppose that the velocity \(v(t)\) (in meters per second) of a sky diver falling near the Earth's surface is given by the following exponential function, where time \(t\) is the time after diving measured in seconds: \(v(t)=55-55 e^{-0.24 t}\).

Step 2 ：We are asked to find how many seconds after diving will the sky diver's velocity be 41 meters per second.

Step 3 ：We can solve this by setting \(v(t)\) equal to 41 and solving for \(t\).

Step 4 ：Doing so gives us the equation \(41 = 55 - 55 e^{-0.24 t}\).

Step 5 ：We can solve this equation for \(t\) using the properties of logarithms.

Step 6 ：First, we subtract 41 from both sides to get \(0 = 14 - 55 e^{-0.24 t}\).

Step 7 ：Then, we divide both sides by -55 to get \(-\frac{14}{55} = e^{-0.24 t}\).

Step 8 ：Taking the natural logarithm of both sides gives us \(\ln(-\frac{14}{55}) = -0.24 t\).

Step 9 ：Finally, we divide both sides by -0.24 to solve for \(t\).

Step 10 ：Doing so gives us \(t = \frac{\ln(-\frac{14}{55})}{-0.24}\).

Step 11 ：Rounding to the nearest tenth, we find that \(t \approx 5.7\).

Step 12 ：Final Answer: The sky diver's velocity will be 41 meters per second approximately \(\boxed{5.7}\) seconds after diving.