# River Rescue problem

1. Sep 7, 2008

### rgalvan2

A child is in danger of drowning in the Merimac river. The Merimac river has a current of 3.1 km/hr to the east. The child is 0.6 km from the shore and 2.5 km upstream from the dock. A rescue boat with speed 24.8 km/hr (with respect to the water) sets off from the dock at the optimum angle to reach the child as fast as possible. How far from the dock does the boat reach the child?

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/fall08/homework/02/IE_rescue/boat.gif [Broken]

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/fall08/homework/02/IE_rescue/help/help/help/help_t/pic6.gif [Broken]

The hardest calculation in this problem is to determine the time it takes for the swimmer to be rescued (and thus the time the dock is moving).
What is t?

So I need to use Pythagorean Theorem to find z. I found it to be 2.57. Then I used d=vt to find the time it takes for the swimmer to be rescued which is the time the dock is moving. So for v, the problem says to use the boat's velocity which is 24.8. I get t=.104 hr. It keeps saying I'm wrong. What am i doing wrong? Help Please!!!

Last edited by a moderator: May 3, 2017
2. Sep 8, 2008

### HallsofIvy

Staff Emeritus
Why are you treating the dock as moving? The child is not motionless. The child is also floating down river at 3.1 km/hr. I think this problem is very simple if you set your frame of reference in the river. If you are entering the solution into a computer, are you expected to give the time in hours or minutes?

3. Sep 8, 2008

### rgalvan2

The help links are making me use the water/child as the reference frame. The time is entered in hours.

4. Sep 8, 2008

### mezarashi

Can you clarify the question again? Are you looking for distance from the dock or time?
In my understanding of the question, when the boat is released from the dock, the boat is stationary with respect to the child. Then it's a matter of just directly driving to the child.

5. Sep 8, 2008

### LowlyPion

What distance (d from your figure) did you calculate? Is that what's asked?

Last edited by a moderator: May 3, 2017
6. Sep 8, 2008

### rgalvan2

I am trying to solve for d. This is just a help link, it just walks me through what I need to do. This help section is having me solve for z and using that in the equation d=vt as the distance. Since we are using the boat's speed, we have v and the distance z. I solved for t giving me .1 hours. This specific help tells me to find z by Pythagorean Theorem which I did: $$\sqrt{2.52 + .62}$$ = 2.57=z. So I plugged z into d=vt: 2.57km=24.8km/hr(t)
t=2.57km/24.8km/hr=.1hr. The computer is not accepting my answer. Am I making a stupid mistake or what?

7. Sep 8, 2008

### LowlyPion

I understand. I get that 2.57 is the distance through the water. But isn't the answer they seek the distance from the dock at the time of rescue - at .1 hr? At a time when the x-leg is .1 hr shorter = .31 km shorter?