A motorboat travels 9 miles downstream (with the current) in 30 minutes. The return trip upstream (against the current) takes 90 minutes.
Which system of equations can be used to find $x$, the speed of the boat in miles per hour, and $y$, the speed of the current in miles per hour? Recall the formula $d=r t$.
A. [tex]
\begin{array}{l}
9=0.5(x-y) \\
9=1.5(x+y)
\end{array}
[/tex]
B. [tex]
\begin{array}{l}
9=1.5(x-y) \\
9=0.5(x+y)
\end{array}
[/tex]
C. [tex]$0.5=9(x-y)$\\
$1.5=9(x+y)$[/tex]
D. [tex]$1.5=9(x-y)$\\
$0.5=9(x+y)$[/tex]
Answer (1)
Define x as the speed of the boat and y as the speed of the current.
Downstream: 9 = 0.5 ( x + y ) .
Upstream: 9 = 1.5 ( x − y ) .
The system of equations is 9 = 1.5 ( x − y ) 9 = 0.5 ( x + y ) .
Explanation
Setting up the equations Let x be the speed of the boat in miles per hour and y be the speed of the current in miles per hour. When traveling downstream, the speed of the boat is x + y . The distance traveled downstream is 9 miles and the time is 30 minutes, which is 0.5 hours. Using d = r t , we have 9 = 0.5 ( x + y ) . When traveling upstream, the speed of the boat is x − y . The distance traveled upstream is 9 miles and the time is 90 minutes, which is 1.5 hours. Using d = r t , we have 9 = 1.5 ( x − y ) . Therefore, the system of equations is 9 = 0.5 ( x + y ) and 9 = 1.5 ( x − y ) .
Examples
Understanding how boats move in currents helps in many real-world scenarios. For example, cargo ships need to account for ocean currents to optimize their routes and save fuel. Similarly, kayakers or canoeists need to understand how river currents affect their speed and direction to navigate effectively. This problem demonstrates how to translate a real-world situation into a mathematical model, which is a crucial skill in engineering and physics.
