abrk
A child is in danger of drowning in the Merimac river. The Mericam river has a current of 3.1 k/h to the east. The child is 0.6 km from the shore and 2.5 k upstream from the dock. A rescue boat with speed 24,8 km/h (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?

Umer.banday
dude...i did this by taking the vel of boat wid respect to ground...
D___________<2.5km>_________A

................................................................,
!_______________________________/
!_____^_______________________/_!
!___0.6km_________________/_\$_!
!____v___________________/______!
!______________________/________!
!......................................../....................!

C____<----x---->______E <2.5-x>B

Now point A is the dock..C is the initial position of child..E is the position of child at time t ..... \$ is the angle...now velocity of boat wrt ground (Vbg) is equal to vel of boat wrt river(Vbr) plus vel of river wrt ground(Vrg)..or (Vbg)=(Vbr)+(Vrg)..equation 1..Now (Vbr) = -248 (sin \$) î -248(cos \$)j and (Vrg) = -3.1î ..So using these in equatiön 1, we get (Vbg) = [-(248 sin \$ + 3.1)î -248c Cos \$ j].eqn 2.now as the baby flows with water,its vel is same as water..so 3.1=x/t...eqn 3..now from eqn 2, x component of vel of the boat =2.5-x/t..or 248sin\$ = (2.5-x)/t -3.1 ..eqn 4..now for y component, 248cos\$ =0.6/t ..eqn 5....now divide eqn 4 by 5 and use x=3.1t and tan\$ = (2.5-x)/0.6 to solve this and find x...enjoy..Note:The underscores in the fig r used in place of space...
Umer.banday
D < 2.5 km > A
..................................................................
! /
! ^ /_!
! 0.6 km / \$ !
! v / !
! / !
!.........................................../..................!
C <----x----> E <2.5-x>B
abrk
i'm not that bad but i can't understand a thing of what you've written...
Guest
Dude...u had given that 248 is the velocity of boat with respect to water...first of all assume that the boat comes at any general angle \$ ..and break the velocity into its components...now in order to find the velocity of boat with respect to ground , we have to add vectorially the velocity of river with respect to ground...once you do that you get Vbg in component form then...write the equations of motion for the baby and the boat..in case of boat ,you will have 2 equations..one for x component and one for y component..solve these three equations to get x..then ur answer is 2.5-x
PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.