1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

River Rescue ( with attachement )

  1. Apr 11, 2005 #1
    Q. 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?

    I know in this problem we need to use Pythagoras Theorem to find one of the sides, actually I think there would be two triangles....

    But I am getting stuck somewhere,

    Anysights or better ideas to solve this problem are appreciated.


    Attached Files:

  2. jcsd
  3. Apr 11, 2005 #2
    Yeah, you got a pythagoreean theorem problem here.
    1. Two of the sides are given to you already. 2.5km for the dock to the boy and the boy is .6km from the shore (I'm assuming the same side as the dock) You should be able to solve a^2 + b^2 = c^2 with those two variables. Note that the boat is moving in relation to the water. Although the river is moving at 3.1m/s the boys motion is also in relation to the water in this problem so he has a relative speed of 0. So don't worry about their motion to the shore just yet.
    2. Once you have the length of side c you know how far the boat must travel to reach the boy. You need to know how much time the boat took to get to him. v=d/t, d=vt, t=d/v
    3. Notice the question asks how far away they are from the dock. The dock moves 3.1 seconds upstream in relation to the water that is rushing past it. That will make it closer to the boat and the boy. For every second the boat was moving the dock was moving 3.1 seconds in relation to the water towards them. d=vt. Multiply the docks relative speed by the total time of the boats movement in the water. This gives you the horizontal distance moved by the dock (assuming the rivers motion is on the x axis and the boy's distance is the y axis)
    4. You now have another pythagorean theorem to figure out. Subtract the distance you got from #3 from the starting x distance of 2.5km. This will take into account the rivers speed and give you a new x distance that represents the boats distance on the x axis from the dock. The y axis is unchanged.
    5. Use the pythagorean theorem to find the distance of the boat from the dock and you will have your answer.

    I'm not the best at math so you might want to verify this with someone else. I just saw that nobody helped you with this simple question. They must all have been asleep or watching the supervolcano on discovery channel. Maybe try to finish your homework before sunday night at midnight next time. :wink: I still have homework to do too.

    Anyway I hope that helped, but you should know that I happen to live right by the merrimack river. I can look out the window and see it and I don't see any drowning boys out there right now so there is really no need to rescue anyone.

    What was the question?
  4. Apr 11, 2005 #3
    Merrimack river is in Massachussetts.... rite
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: River Rescue ( with attachement )
  1. River Rescue (Replies: 0)

  2. Space Rescue Program (Replies: 0)

  3. River Rescue problem (Replies: 6)

  4. Rescue Plane Question (Replies: 3)

  5. Alaskan Rescue Plane (Replies: 16)