Question: The swim portion of a triathlon takes place in a river and is an out-and-back course. The river has a steady current that is 0.400m/s and the athletes swim against the current for 750m before turning and swimming with the current for the second 750m. If one competitor can swim at a steady speed of 1.10m/s in still water, how long will their swim take in this river? Compare this to how long it would take in the absence of any current.
My question: I know that for this problem I have to add the relative velocities and subtract the relative velocities when the swimmer is going in the opposite direction, but I am not exactly sure where to start or if there is a specific equation I should use?
(Vs-Vw)t1 = 750 m.
(1.10-0.4)t = 750
0.7t1 = 750
t = 1071.4 s. = 17.9 min. = Time to swim
upstream.
(1.10+0.4)t2 = 750 m.
1.5t2 = 750
t2 = 500 s. = 8.33 min. = Time to swim
downstream.
t1+t2 = 17.9 + 8.33 = 26.2 min. = Time to swim upstream and back.
t3 = d/V = (750+750)/1.1 = 1363.6 s. =
22.7 min., No wind.
26.2-22.7 = 3.47 min. less with zero wind speed.
To solve this problem, you can use the concept of relative velocity and apply it to the swimmer's speed and the river's current. First, let's understand the situation and break it down into two parts:
Part 1: Swimming against the current
In this part, the swimmer is swimming against the river's current, so the current will act as an opposing force, decreasing the swimmer's effective speed.
To find the effective speed of the swimmer when swimming against the current, subtract the current's velocity from the swimmer's velocity:
Effective speed = Swim speed - Current speed
In this case, the swimmer's speed is given as 1.10 m/s and the current speed is 0.400 m/s. So the effective speed of the swimmer when swimming against the current is:
Effective speed = 1.10 m/s - 0.400 m/s = 0.700 m/s
Now, using this effective speed, you can calculate the time it takes the swimmer to cover the first 750m:
Time = Distance / Speed
Time (against current) = 750m / 0.700 m/s = 1071.43 s
Part 2: Swimming with the current
In this part, the swimmer is swimming with the river's current, so the current will act as an assisting force, increasing the swimmer's effective speed.
To find the effective speed of the swimmer when swimming with the current, add the current's velocity to the swimmer's velocity:
Effective speed = Swim speed + Current speed
In this case, the swimmer's speed is still 1.10 m/s, but now the current speed is acting in the same direction, so it becomes positive. Therefore, the effective speed of the swimmer when swimming with the current is:
Effective speed = 1.10 m/s + 0.400 m/s = 1.50 m/s
Using this effective speed, you can calculate the time it takes the swimmer to cover the second 750m:
Time = Distance / Speed
Time (with current) = 750m / 1.50 m/s = 500 s
Total Time taken
To find the total time taken for the swim, you need to add the times taken for swimming against and with the current:
Total Time = Time (against current) + Time (with current)
Total Time = 1071.43 s + 500 s = 1571.43 s
In the absence of any current, the swimmer would swim at a constant speed of 1.10 m/s throughout the entire distance of 1500m. To calculate the time it would take in the absence of any current, you can use the formula:
Time = Distance / Speed
Time (no current) = 1500m / 1.10 m/s ≈ 1363.64 s
Therefore, the swim will take approximately 1571.43 seconds with the current and 1363.64 seconds in the absence of any current.