Speed of river =v Speed of man in still water =2v ∴ Speed upstream =(2v−v) Speed downstream =(2v+v) Now, according to the question, total time taken in going upstream and back downstream to the starting point =t ∴tup +tdown =t⋅⋅⋅⋅⋅⋅⋅(i) Let the total distance on one way =d∴tup =
d
2v−v
⇒tdown =
d
2v+v
Putting these values is Eq. (i) we get,
d
2v−v
+
d
2v+v
=t ⇒d[
1
v
+
1
3v
]=t⇒
d
v
[
3+1
3
]=t ⇒t=
4d
3v
⋅⋅⋅⋅⋅⋅⋅(ii) It is given that time taken by the man to cover the same distance in still water =t0 ⇒t0=