# Homework Help: Please, help! River Current

1. May 23, 2004

### PoetryInMotion

Okay, this may sound silly... but it's driving me crazy! Some time ago, my friend showed me an extra credit problem for her class and ever since I haven't been able to figure it out. Now classes are over and we've all moved home, but it's still bugging me to no end. So, if you can help... please do!

A boat starts traveling down a river on a boat at point A and at point B a hat falls out of the boat (and floats), but the driver doesn't realize that the hat fell out until he reaches point C. The man turns around the boat and meets up with the hat at point D.

The distance between point A and B is 1 mile.
The time it takes to travel from point B to C is 20 minutes.
It takes the same amount of time for the boat to travel to point C to D as it takes the hat to travel from point B to point D.
The boat is traveling at a constant speed.

What is the current of the water?

And here's a little diagram of the picture her teacher drew them.

B

A D

Help!

2. May 24, 2004

### HallsofIvy

"It takes the same amount of time for the boat to travel to point C to D as it takes the hat to travel from point B to point D."

Surely that's not true. The hat didn't sit at point B waiting for 20 minutes until the boat reached point C and then start floating downstream as soon as the boat turned back upstream! What is true is that the time taken for the boat to go down stream from B to C and then back from C to D is the same as the time taken for the hat to float from B to D. The boat and hat were together at point B and then again at point D. The time elapsed is the same for both.

Call the speed of the boat, relative to the water, u and the current speed s. Since the boat is initially traveling down stream, its speed, relative to the bank, is u+ v. When it turns and goes back up stream, its speed, relative to the bank, is u- v. Of course, the speed of the hat, relative to the bank, once it has fallen out of the boat is the same as the speed of the water: v.

When the hat falls out of the boat, the boat continues downstream for an unknown time t1 at speed u+v to point C: it travels a distance (u+v)t1. It then turns around and travels upstream for another unknown time t2 at speed u-v to point D: it travels a distance (u-v)t2. The actual distance from point B to point D is (u+v)t1- (u-v)t2= u(t1- t2)+ v(t1+ t2).

During that time, the hat is floating downstream at speed v for time t1+ t2. The distance it floats, which is also the distance from B to D is v(t1+ t2.

Setting those two different calculations for the distance from B to D equal:
u(t1- t2)+ v(t1+ t2)= v(t1+ t2.
Notice that the "v(t1+ t2)" terms on each side cancel leaving us with u(t1- t2)= 0 or t1= t2. The time going downstream must be equal to the time going upstream!

We can now note that t1 is given as 20 minutes so t2 is also 20 minutes.

Unfortunately, we still have no way to find either u or v separately. The information that "The distance between point A and B is 1 mile" doesn't help at all since that has nothing to do with the hat.

If we were told "the distance between point B and point D" is 1 mile, then, since we know it took the boat 20 minutes to go from point B to point C, and that is the same as the time the to go back up from C to D, we would know that the hat was floating for 40 minutes. We could then calculate that v= 1 mile/40 minutes= 1.5 miles per hour.