A river 500m wide flows at a rate of 4km/h. A swimmer who can swim at 8km/h in still water, wishes to cross the river straight.
(i) Along what direction must be strike?
(ii) what should be his resultant velocity?
(iii) what is the time taken for crossing the river?
pls explain the answer with a reference diagram!
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Answer:
Solution :
(i) Refer to fig. 2(c ) 22,
velocity of river =4kmh-1=(OA)
Velcity of swinmmer in still water
vs0=8kmh-10=(OB)
The swimmer will cross the river straitght if the resultant velocity
→
v
of
→
v
rand
→
v
s is perpendiculat to the bank fof the river, i.e., along (
→
O
C). This will be possible if the swimmer groes upatream of the river along OB, making an angle θ with OC. In right ΔOCB, sin θ
NC
OB
=
vr
vs0=
4
8
=
1
2
=sin3000
:. θ=3000 (ii) Resultant velocity of the swimmer
v=
√
v
2
s
-v
2
r
=
√
820-42=
√
48
=4
√
3
kmh-1=4
√
3
×5/18=1.92m/s(iii)Timetaken→crosstheriver,
t=(width of river)/v =(500 m)/(1.92 ms^(-1)) =260.4 s`.
.
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