Suppose that the velocity $v(t)$ (in $\mathrm{m} / \mathrm{s})$ of a sky diver falling near the Earth's surface is given by the following function, where time $t$ is measured in seconds.
\[
v(t)=57\left(1-e^{-0.25 t}\right)
\]
Find the initial velocity of the sky diver and the velocity after 6 seconds. Round your answers to the nearest whole number as necessary.
Initial velocity:
\[
\prod \mathrm{m} / \mathrm{s}
\]
Velocity after 6 seconds:
$\mathrm{m} / \mathrm{s}$
Answer
So, the velocity after 6 seconds is \(\boxed{44}\) m/s.
Step 1 ：The initial velocity of the skydiver can be found by substituting \(t=0\) into the velocity function.
Step 2 ：\[v(0)=57(1-e^{-0.25 \times 0})\]
Step 3 ：So, the initial velocity of the skydiver is \(\boxed{0}\) m/s.
Step 4 ：The velocity after 6 seconds can be found by substituting \(t=6\) into the velocity function.
Step 5 ：\[v(6)=57(1-e^{-0.25 \times 6})\]
Step 6 ：So, the velocity after 6 seconds is \(\boxed{44}\) m/s.
