 # A swimmer can swim with velocity v wrt still water in a river flowing with velocity u in the same direction as man. There is a float moving with river. The man overtakes the float and gets a lead of l and then returns back to the float so the time taken in this process is what?The answer is2l/v Plz explain how??

11 years ago
1. Relative speed of man=u+v (downstream direction) Distance=l Relative speed of man and float=u+v-u (same direction)=v T1=l/t 2. Relative speed of man=v-u(upstream) Distance =l Relative speed of man and float=v-u+u (opposite direction)=v T2=l/v Therefor total t is l/v + l/v =2l/v
11 years ago First consider the case where the swimmer gains a lead of "l" to the boat.

to find the time required for the swimmer to reach P, we need to know the relative velocity of the swimmer ...

here he is going downstream therefore,velocity of swimmer V(s)=v+u

velocity of boat V(b)=u

therefore the relative velocity of the swimmer with respect to the boat ,

V=(v+u) - u

=v

there fore time taken,t(1)=l/v

similarly in the upstream the relative velocity of the swimmer is V=(v-u) - (-u)

=v - u + u

=v

there fore the time taken to return the boat t(2)=l/v

there fore the total time taken T=t(1)+t(2)=l/v + l/v = 2l/v

pls approve...

BEST OF LUCK

11 years ago

hi

relative velocity of man (down stream)=v+u.

speed of boat =u

therefore the velocity of man wrt boat is v+u-u=v

time1=l/v

time2(to return)=l/v

therefore total time is 2l/v

11 years ago

downstream ,velocity of swimmer V(s)=v+u

velocity of boat V(b)=u

therefore the relative velocity of the swimmer with respect to the boat , V=(v+u) - u =v

time taken,t(1)=l/v

similarly in the upstream the relative velocity of the swimmer is V=(v-u) - (-u) =v - u + u =v

then the time taken to return the boat t(2)=l/v

the total time taken:

T=t(1)+t(2)=l/v + l/v = 2l/v

All the best! 