2 swimmers start from point A on one bank of a river to reach point B on the other bank, lying directly opposite
to point A. One of them crosses the river along the straight line AB, while the other swims at right angles to the stream and
then walks the distance with he has been carried away by the stream to get to point B. Velocity of each swimmer in still
water is 2.5 km hr-1 and the stream velocity is 2km h-1. Width of river is 300m
Q: Calculate time required by each swimmer to move from A to B in hour
Answers
Answer:
Let one of the swimmer (say 1) crosses the river along AB, which is obviously the shortest path. Time taken to cross the river by the swimmer 1.
t
1
=
v
′
2
−v
0
2
d
, (where AB=d is the width of the river) (1)
For the other swimmer (say 2), which follows the quickest-path, the time taken to cross the river.
t
2
=
v
′
d
(2)
In the time t
2
, drifting of the swimmer 2, becomes
x=v
0
t
2
=
v
′
v
0
d, (using equation 2) (3)
If t
3
be the time for swimmer 2 to walk the distance x to come from C to B (shown in figure above), then
t
3
=
u
x
=
v
′
u
v
0
d
(using equation 3) (4)
According to the problem t
1
=t
2
+t
3
or,
v
′
2
−v
0
2
d
=
v
′
d
+
v
′
u
v
0
d
On solving, we get
u=
(
v
′2
1−v
0
2
)
−
2
1
−1
v
0
=3.0km/h