A motorboat can maintain a constant speed of 31 miles per hour relative to the water. The boat makes a trip upstream to a certain point in 32 ​minutes; the return trip takes 30 minutes. What is the speed of the​ current?

Let \( v_b \) be the speed of the motorboat relative to the water, which is 31 miles per hour, and let \( v_c \) be the speed of the current.

### Step 1: Convert the time into hours
- Time upstream: \( 32 \) minutes \( = \frac{32}{60} = \frac{8}{15} \) hours.
- Time downstream: \( 30 \) minutes \( = \frac{30}{60} = \frac{1}{2} \) hours.

### Step 2: Define the effective speeds
- Upstream, the speed of the boat relative to the ground is \( v_b - v_c \), because the current opposes the boat.
- Downstream, the speed of the boat relative to the ground is \( v_b + v_c \), because the current helps the boat.

### Step 3: Set up distance equations
Let \( d \) represent the distance to the certain point. Using the formula \( \text{distance} = \text{speed} \times \text{time} \), we have two equations:

1. For the upstream trip: 
\[
d = (v_b - v_c) \times \frac{8}{15}
\]
2. For the downstream trip:
\[
d = (v_b + v_c) \times \frac{1}{2}
\]

Since the distance \( d \) is the same for both trips, set the two equations equal to each other:

\[
(v_b - v_c) \times \frac{8}{15} = (v_b + v_c) \times \frac{1}{2}
\]

### Step 4: Solve for \( v_c \)
Substitute \( v_b = 31 \) miles per hour:

\[
(31 - v_c) \times \frac{8}{15} = (31 + v_c) \times \frac{1}{2}
\]

Multiply both sides by 30 to eliminate the fractions:

\[
(31 - v_c) \times 16 = (31 + v_c) \times 15
\]

Expand both sides:

\[
496 - 16v_c = 465 + 15v_c
\]

Move terms involving \( v_c \) to one side:

\[
496 - 465 = 16v_c + 15v_c
\]

\[
31 = 31v_c
\]

Thus, \( v_c = 1 \) mile per hour.

### Conclusion
The speed of the current is \( 1 \) mile per hour.

Would you like any more details or clarifications?

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Here are some related questions to think about:
1. How would the trip time change if the current speed increased by 50%?
2. What would happen if the boat could no longer maintain a constant speed?
3. How would the time taken downstream change if the current reversed direction?
4. How can you modify this problem for distances in kilometers instead of miles?
5. What if the current speed varied over the trip?

**Tip:** Always break down problems involving motion into speed, distance, and time for clarity.

A motorboat can maintain a constant speed of 31 miles per hour relative to the water. The boat makes a trip upstream to a certain point in 32 minutes; the return trip takes 30 minutes. What is the speed of the​ current?

Solve the problem of determining the speed of the current when a motorboat travels upstream at 31 mph, taking 32 minutes upstream and 30 minutes downstream. Learn how to apply the distance-speed-time relationship and solve linear equations. Suitable for students in Grades 9-11.

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