https://file.aiquickdraw.com/mathbot/file/17299850453715kw39trb.jpg

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.

On the interval [4,15], solve the last 4 parts of the problem involving velocity and distance.

Solve a calculus problem involving velocity and distance analysis on the interval [4,15]. This problem requires finding critical points, analyzing maximum and minimum distances, and calculating when velocity is zero. Suitable for high school or college-level calculus students.

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