Let's analyze each part of the remaining questions and solve them.

The given function is: $d(t) = \frac{100}{5 + 4 \sin(t)}$ and the velocity function $v(t)$ (already derived in part (c)) is: $v(t) = -\frac{400 \cos(t)}{\left(5 + 4 \sin(t)\right)^2}$

Part (f): First time when $v(t) = 0$ on the interval $[4, 15]$

To find when the velocity $v(t)$ is zero, set: $v(t) = -\frac{400 \cos(t)}{\left(5 + 4 \sin(t)\right)^2} = 0$ Since the fraction equals zero when the numerator is zero, we need: $\cos(t) = 0$ This occurs at: $t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$ Converting these into decimal approximations within $[4, 15]$:

• $\frac{3\pi}{2} \approx 4.71$ (this is the first solution in the interval)
• $\frac{7\pi}{2} \approx 10.99$
• $\frac{11\pi}{2} \approx 17.28$ (out of the interval)

So, the first time $t$ when $v(t) = 0$ in $[4, 15]$ is: $t = \frac{3\pi}{2} \approx 4.71$

Part (g): Last time when $v(t) = 0$ on the interval $[4, 15]$

From the above solutions within $[4, 15]$:

• The last time $t$ when $v(t) = 0$ is $t = \frac{7\pi}{2} \approx 10.99$.

Part (h): Maximum distance from the origin on $[4, 15]$

To find the maximum distance, evaluate $d(t)$ at critical points (where $v(t) = 0$) and endpoints of the interval.

1. Endpoints:

   • $d(4) = \frac{100}{5 + 4 \sin(4)} \approx 11.68$
   • $d(15) = \frac{100}{5 + 4 \sin(15)} \approx 15.59$

2. Critical points (where $v(t) = 0$):

   • $d\left(\frac{3\pi}{2}\right) = \frac{100}{5 + 4 \sin\left(\frac{3\pi}{2}\right)} = \frac{100}{5 - 4} = 100$
   • $d\left(\frac{7\pi}{2}\right) = \frac{100}{5 + 4 \sin\left(\frac{7\pi}{2}\right)} = \frac{100}{5 - 4} = 100$

Thus, the maximum distance is: $100$

Part (i): Minimum distance from the origin on $[4, 15]$

Again, use the same points.

1. Endpoints:

   • $d(4) \approx 11.68$
   • $d(15) \approx 15.59$

2. Critical points:

   • $d\left(\frac{3\pi}{2}\right) = 100$
   • $d\left(\frac{7\pi}{2}\right) = 100$

The smallest value is at $d(4) \approx 11.68$.

So, the minimum distance is approximately: $11.68$

Summary of Answers:

• (f) First time when $v(t) = 0$: $t \approx 4.71$
• (g) Last time when $v(t) = 0$: $t \approx 10.99$
• (h) Maximum distance: $100$
• (i) Minimum distance: $\approx 11.68$

Would you like more details on any step or have questions about the calculations?

Further Related Questions:

1. How would we approach finding minimum or maximum points if $d(t)$ were more complex?
2. How does average velocity differ from instantaneous velocity in such problems?
3. Could we approximate these values without a calculator?
4. How would the answers change if the interval were different?
5. What does it mean physically when $v(t) = 0$?

Tip: When solving optimization problems, always evaluate at critical points and endpoints for a closed interval.