On a Boat

SOLUTION 1

One general strategy for solving this kind of problem is to look at an extreme case of the changing variable. In this case, consider a current that is almost capable of pushing the boat backwards. If the current is strong enough, the boat's speed can be reduced to as slow a crawl as you can imagine. And it will take the boat an eternity to go from B back to A. The amount of time spent going from A to B might be reduced to next to nothing, but infinity plus epsilon is still infinity of course!

However, you might be thinking that if the current is smaller, this logic doesn't hold. In fact it does hold, it's just harder to intuitively see what happens! But here's an explanation of why the journey will be longer, even if the current is quite minimal:

SOLUTION 2

Let the speed of the boat in still water be $S$ meters/second, and let the current be $C$ meters/second.
The effective speed of the boat moving with the current is therefore $S+C$ .
The effective speed of the boat moving against the current is therefore $S-C$ .

Consider a boat traveling at speed $S$ meters per second for 2 seconds . Since distance = rate x time, it will move 2S meters in 2 seconds. However, if, for half of that time (one second), an external force causes it to move at a rate of $S+C$ , then, the boat will instead move $S+C$ meters in tone second. And if, for half of that time, an external force causes it to move at a rate of $S-C$ , then, it will move $S-C$ , meters in that second second. So, if the same force acts both in concert with, and then against the boats motion for equal amounts of time, then the effects of the force cancel out: $(S+C) + (S-C) = 2S$ .

If, however, that same force were to act against the boat for longer than it helped the boat, then the effect would be to net-lengthen the whole trip, right? This latter case is what applies to our problem since, because the current slows the boat down, the boat will be moving against the current for longer than it will be moving with the current. So, regardless of the strength or direction of the current, the entire trip will be longer.

We have a boat that travels at a speed $S$ and the river current flows at speed $C$ . We will assume both speeds are positive and that $C<S$ (otherwise, the boat would not be able to reach the starting point again.) Physics tells us distance is the product of speed and time: $d=st$ , which implies $t=\frac{d}{s}$

Situation 1: No current

If there's no current, the time for both legs of the journey are the same:

$t_{1}=\frac{d}{S}+\frac{d}{S}=\frac{2d}{S}$

Situation 2: Current

Now both legs of the journey do not take the same time.

$t_{2}=\frac{d}{S+C}+\frac{d}{S-C}=\frac{d(S-C+S+C)}{S^{2}-C^{2}}=\frac{2dS}{S^{2}-C^{2}}$

Comparison

To compare more easily, we will make the denominators the same.

$t_{1}=\frac{2d}{S}=\frac{2d(S^{2}-C^{2})}{S(S^{2}-C^{2})}$

$t_{2}=\frac{2dS}{S^{2}-C^{2}}=\frac{2dS^{2}}{S(S^{2}-C^{2})}$

Everything else being equal, the comparison now is between $(S^{2}-C^{2})$ and $S^{2}$ . Since both $S$ and $C$ are positive and $C<S$ , $(S^{2}-C^{2})$ must be smaller, so $t_{1}$ is the smaller value. It will take less time to make the trip if there is no current than if there is current, given a constant boat speed.

I essentially just took a simple-to-calculate case.

I started off by assuming that, without a current, the boat travels the same speed both ways, so it takes half an hour to get from one side to the other, without a current.

Then, I took a case where the current was enough to half the time traveling downstream, so the current is the same as the boat's speed. (We can set the boat's speed to whatever we want, as it isn't limited, so, if you assume it has a speed of 1, D=RT becomes D=T. And, since the distance doesn't change with current, we can set up a new equation with a new speed S that factors in the current, so D=S( $\frac{T}{2}$ ). Finally, we can set them equal, so T=S( $\frac{T}{2}$ ). Solving for S, we get S=2=1+1=R+R, so the current's force is equal to R.)

Then, going the other way, the current is acting against the boat, so subtract the current's force from the speed of the boat to get the time M it takes to get back. D=(R-R)M=(0)M=0

In this case, not only does the boat not make it back within an hour, it doesn't make it back at all!

(It should be noted that acceleration isn't always uniform, and if it were a rowboat, the sailor might make it a little bit into the water before getting pushed back, because his initial speed is greater than his average speed for the whole trip, and greater than the current's force)

Very simple, no calculation needed. the boost that the boat gets from the current shortens the length of time that it moves at the faster rate of speed. The boat spends more time fighting the current, and thus the total time is longer, not the same, or shorter. Of course viscosity changes with the velocity of the boat, but I assume that effect was considered negligible for the purpose of the problem