v B = λ /2 τ = 3.0 km per hour

Sol. At t = 0, raft (a float of timber) and motor boat are at point A. The velocity of raft is equal to velocity of stream.

At t = τ = 60 min, the motor boat is at point P and raft is at point B.

∴ The time taken by raft to reach from A to B = the time taken by motor boat to reach at P from A.

At t = τ + t 0 , both meet at point C,

So, the time taken by raft to reach at C from B is equal to the time taken by the motor boat to reach at C from P in upstream motion. This time is equal to t 0 .

Let

v A = actual velocity of motor boat,

v B = actual velocity of stream = velocity of raft

∴ During down stream,

v c = relative velocity of motor with respect to stream.

∴ v 0 = v A – v B

∴ v A = v 0 + v B

∴ τ = =

But AB = distance travelled by raft in time τ = v B τ

During upstream,

v 0 = v A + v B

∴ v A = v 0 – v B

∴ PC = distance travelled by motor boat in upstream in time t 0 = (v 0 – v B ) t 0

BC = distance travelled by raft in time t 0 = v B t 0

According to fig.

 AP – PC = AC = λ

or (v 0 + v B ) τ – (v 0 – v B ) t 0 = λ

or v 0 t +v B t – v 0 t 0 + v B t 0 = λ ..........(i)

Also,  AB + BC = λ

or v B t + v B t 0 = λ

or v B = ..........(ii)

From equation (i) and (ii) we get

v 0 t +v B t – v 0 t 0 + v B t 0 = λ

or

or v 0 τ 2 + τ λ – v 0 t 0 2 + λ t 0 = λ ( τ + t 0 )

or v 0 τ 2 – v 0 t 0 2 = λ ( τ + t 0 ) – τ λ – t 0 λ

or τ = t 0

From (i) we have

v 0 τ +v B τ – v 0 t 0 + v B t 0 = λ

putting τ = t 0

we get, v B = .