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.