Optimistic Math Problem Solving: Tim's Kayaking Adventure

How far could Tim have paddled in still water during his kayaking trip?

Given data: Tim paddled his kayak 12 km upstream against a 3 km/h current and back again in 5 hours and 20 minutes. In that time, how far could he have paddled in still water?

Answer:

In still water, Tim could have paddled a distance of 24 km.

Tim paddled his kayak 12 km upstream against a 3 km/h current. This means his effective speed relative to the ground was his paddling speed minus the current speed, which is the speed of the kayak in still water minus the speed of the current: v₁ = v - c = (v - 3) km/h.

Next, Tim paddled back downstream the same distance of 12 km. This time, his effective speed relative to the ground was his paddling speed plus the current speed: v₂ = v + c = (v + 3) km/h.

The total time for the round trip was given as 5 hours and 20 minutes, which is equivalent to 5.33 hours. Since the time it takes to travel a certain distance is equal to the distance divided by the speed, we can set up the following equation based on the given information:

12/(v - 3) + 12/(v + 3) = 5.33

To solve this equation and find the value of v (the speed of the kayak in still water), we can use algebraic techniques such as finding a common denominator, simplifying the equation, and then solving for v. However, in this case, the exact solution involves more complex calculations and decimals. Therefore, we'll use an approximation method to find the approximate value of v.

After solving the equation, we find that the approximate value of v is 5.5 km/h. Now, we can calculate the distance Tim could have paddled in still water by multiplying his approximate speed by the total time of 5.33 hours: 5.5 km/h * 5.33 hours = 29.315 km.

However, we need to consider that Tim paddled 12 km upstream and 12 km downstream in opposite directions. Thus, the total distance he could have paddled in still water is 12 km + 12 km = 24 km.

To know more about algebraic techniques, refer to reliable sources.

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