Two friends went rowing on the river. They first traveled with current for 3 hours. The way back took them 4 hours and 48 minute

Question

tamaranim1

2 months ago

10

Two friends went rowing on the river. They first traveled with current for 3 hours. The way back took them 4 hours and 48 minute

s. Find the speed of the river's current if they can row in still water at a speed of 6.5 mph.

Mathematics

2 answers:

Inessa [12.5K] · Answer 1 · 2026-03-14T08:59:55+03:00

Answer:

The current's speed is 1.5 mph.

Step-by-step explanation:

We need to establish an expression for each journey. For the first part, it's clear the boat is moving alongside the current, which implies their speeds add together due to being in the same direction. Conversely, in the second journey, the speeds are subtracted because the boat is going against the current's direction.

At this point, we know the boat's speed is 6.5 mph, while the current's speed is unknown and will be referred to as c. We also presume the movement is at a constant speed, thus we can utilize the formula: $d = vt$ ; where d represents distance, v indicates speed, and t stands for time.

First movement:

$d_{1} = (6.5 + c)(3hr)$

Second movement:

$d_{2} = (6.5 - c)(4.8hr)$

The value 4.8 is the converted figure, as the problem specifies 4 hours and 48 minutes, which we need to convert the minutes to hours. Since we know that 1hr equals 60 min:

$48min\frac{1hr}{60min}=0.8hr$ ; the time for the second trip equates to 4.8hr.

Now, since the scenario indicates the boat returned, this indicates both trips covered the same distance, allowing us to equate both equations, then isolate c to determine the current's speed:

$d_{1}=d_{2} \\(6.5 + c)(3hr)=(6.5 - c)(4.8hr)\\19.5 + 3c = 31.2-4.8c\\4.8c + 3c=31.2-19.5\\7.8c=11.7\\c=\frac{11.7}{7.8}=1.5mph$