Question

A motorboat motorboat can maintain a constant speed of 20 miles per hour relative to the water in going 21 miles upstream and then returning. The total time for the trip was 2.5 hours?

Answer 1

To solve this problem, we will first establish the key variables:

1. Let \( v_b = 20 \) miles per hour (the speed of the motorboat relative to the water).
2. Let \( v_c \) be the speed of the current (in miles per hour).
3. The distance traveled upstream is \( d = 21 \) miles and the same distance downstream.

### Trip Time Calculations

1. **Upstream Time**:
When going upstream, the effective speed of the motorboat is reduced by the speed of the current:
\[
\text{Speed upstream} = v_b - v_c = 20 - v_c \, \text{miles/hour}.
\]
The time taken to go upstream is:
\[
t_{\text{up}} = \frac{d}{\text{Speed upstream}} = \frac{21}{20 - v_c} \, \text{hours}.
\]

2. **Downstream Time**:
When going downstream, the effective speed of the motorboat is increased by the speed of the current:
\[
\text{Speed downstream} = v_b + v_c = 20 + v_c \, \text{miles/hour}.
\]
The time taken to go downstream is:
\[
t_{\text{down}} = \frac{d}{\text{Speed downstream}} = \frac{21}{20 + v_c} \, \text{hours}.
\]

### Total Time for the Trip

According to the problem, the total time for the trip is 2.5 hours:
\[
t_{\text{up}} + t_{\text{down}} = 2.5.
\]
Substituting our expressions for \( t_{\text{up}} \) and \( t_{\text{down}} \):
\[
\frac{21}{20 - v_c} + \frac{21}{20 + v_c} = 2.5.
\]

### Combining the Fractions

To combine the left side, we'll find a common denominator:
\[
\frac{21(20 + v_c) + 21(20 - v_c)}{(20 - v_c)(20 + v_c)} = 2.5.
\]
Simplifying the numerator:
\[
21(20 + v_c) + 21(20 - v_c) = 420 + 21v_c + 420 - 21v_c = 840.
\]
Now the equation becomes:
\[
\frac{840}{(20 - v_c)(20 + v_c)} = 2.5.
\]

### Cross-Multiplying

Cross-multiply to eliminate the fraction:
\[
840 = 2.5(20 - v_c)(20 + v_c).
\]
Expanding the right side:
\[
840 = 2.5(400 - v_c^2).
\]
Then,
\[
840 = 1000 - 2.5v_c^2.
\]

### Rearranging

Now rearranging gives us:
\[
2.5v_c^2 = 1000 - 840,
\]
\[
2.5v_c^2 = 160.
\]
Dividing by 2.5:
\[
v_c^2 = \frac{160}{2.5} = 64.
\]
Taking the square root gives:
\[
v_c = 8 \, \text{miles/hour}.
\]

### Conclusion

Now that we have the speed of the current, we can summarize:
- Speed of the current: \( v_c = 8 \) miles per hour.
- Speed of the motorboat relative to the water: \( v_b = 20 \) miles per hour.

Thus, the answer is the speed of the current is **8 miles per hour**.