A motor boat can maintain a constant speed of 21 miles per hour relative to the water the boat makes a trip upstream to a certain point in 44 minutes the return trip takes 33 minutes what is the speed of the current 

To solve for the speed of the current, let's define the variables:

- Let \( v_b = 21 \) miles per hour be the speed of the boat relative to the water.
- Let \( v_c \) be the speed of the current (what we are solving for).
- Let \( d \) be the distance from the starting point to the destination (in miles).

When the boat is moving upstream, the effective speed is reduced by the current: \( v_b - v_c \). 
When the boat is moving downstream, the effective speed is increased by the current: \( v_b + v_c \).

### Step 1: Convert the time to hours
- Upstream trip time: \( 44 \, \text{minutes} = \frac{44}{60} = \frac{11}{15} \, \text{hours} \).
- Downstream trip time: \( 33 \, \text{minutes} = \frac{33}{60} = \frac{11}{20} \, \text{hours} \).

### Step 2: Set up the equations for the distance
For the upstream trip:
\[
d = \left(v_b - v_c\right) \times \frac{11}{15}
\]
For the downstream trip:
\[
d = \left(v_b + v_c\right) \times \frac{11}{20}
\]

Since both equations represent the same distance \( d \), we can set them equal to each other:
\[
\left(v_b - v_c\right) \times \frac{11}{15} = \left(v_b + v_c\right) \times \frac{11}{20}
\]

### Step 3: Solve the equation for \( v_c \)
Cancel out the common factor of \( 11 \):
\[
\left(v_b - v_c\right) \times \frac{1}{15} = \left(v_b + v_c\right) \times \frac{1}{20}
\]
Multiply both sides by 60 to eliminate the fractions:
\[
4 \left(v_b - v_c\right) = 3 \left(v_b + v_c\right)
\]
Expand both sides:
\[
4v_b - 4v_c = 3v_b + 3v_c
\]
Collect like terms:
\[
4v_b - 3v_b = 4v_c + 3v_c
\]
Simplify:
\[
v_b = 7v_c
\]

### Step 4: Solve for \( v_c \)
\[
v_c = \frac{v_b}{7} = \frac{21}{7} = 3 \, \text{miles per hour}
\]

Thus, the speed of the current is **3 miles per hour**.

Would you like further explanation on any part of this solution?

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Here are 5 related questions for further practice:

1. If the boat's speed relative to the water were 24 mph, how would the speed of the current change?
2. What would happen to the travel times if the current speed increased by 2 mph?
3. How much longer would the upstream trip take if the current were 5 mph?
4. If the distance were doubled, how would the travel times change with the same current speed?
5. How can you determine the distance traveled if the boat's speed and current speed are both known?

**Tip**: When solving problems involving relative motion, always carefully define the directions and account for how the current (or any external factor) affects the net speed!

A motor boat can maintain a constant speed of 21 miles per hour relative to the water. The boat makes a trip upstream to a certain point in 44 minutes. The return trip takes 33 minutes. What is the speed of the current?

Learn how to calculate the speed of the current using a motor boat's upstream and downstream times. This problem involves key concepts in relative motion, algebra, and the distance-speed-time relationship. Suitable for students in Grades 8-10, with a step-by-step solution provided.

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