HALLP QUICKKKK A tourist sailed against the current on a river for 6 km, and then he sailed in a lake for 15 km. In the lake he sailed for 1 hour longer than he sailed in the river. Knowing that the current of the river is 2 km/hour, find the speed of the boat while it is traveling in the lake.
Accepted Solution
A:
To solve this we are going to use the formula for speed: [tex]S= \frac{d}{t} [/tex] where [tex]S[/tex] is the speed [tex]d[/tex] is the distance [tex]t[/tex] is the time
Let [tex]S_{l}[/tex] be the speed of the boat in the lake, [tex]S_{a}[/tex] the speed of the boat in the river, [tex]t_{l}[/tex] the time of the boat in the lake, and [tex]t_{a}[/tex] the time of the boat in the river.
We know for our problem that the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour: [tex]S_{a}=S_{l}-2[/tex] We also know that in the lake the boat sailed for 1 hour longer than it sailed in the river, so: [tex]t_{l}=t_{a}+1[/tex]
Now, we can set up our equations. Speed of the boat traveling in the river: [tex]S_{a}= \frac{6}{t_{a} } [/tex] But we know that [tex]S_{a}=S_{l}-2[/tex], so: [tex]S_{l}-2= \frac{6}{t_{a} } [/tex] equation (1)
Speed of the boat traveling in the lake: [tex]S_{l}= \frac{15}{t_{l} } [/tex] But we know that [tex]t_{l}=t_{a}+1[/tex], so: [tex]S_{l}= \frac{15}{t_{a}+1} [/tex] equation (2)
Solving for [tex]t_{a}[/tex] in equation (1): [tex]S_{l}-2= \frac{6}{t_{a} } [/tex] [tex]t_{a}= \frac{6}{S_{l}-2} [/tex] equation (3)
Solving for [tex]t_{a}[/tex] in equation (2): [tex]S_{l}= \frac{15}{t_{a}+1} [/tex] [tex]t_{a}+1= \frac{15}{S_{l}} [/tex] [tex]t_{a}=\frac{15}{S_{l}}-1[/tex] [tex]t_{a}= \frac{15-S_{l}}{S_{l}} [/tex] equation (4)