A motor boat sailing against the river flow met a raft, and after a time t1 = 35 minutes it landed on the shore

A motor boat sailing against the river flow met a raft, and after a time t1 = 35 minutes it landed on the shore. Having stood for t2 = 20 minutes, she turned back and overtook the raft in t3 = 40 minutes. Determine the speed of the boat relative to the water υ1, if the speed of the raft is υо = 0.5 m / s.

Given:

t1 = 35 minutes = 2100 seconds – the time interval during which the motor boat reached the shore after meeting the raft;

t2 = 20 minutes = 1200 seconds – the idle time of the motor boat;

t3 = 40 minutes = 2400 seconds – the time after which the boat reached the raft;

v0 = 0.5 m / s – speed of the raft (river flow).

It is required to determine the speed of the motor boat relative to the water v1 (m / s).

After the motor boat met the raft, it covered a distance equal to:

S = (v1 – v0) * t1.

Back she caught up with the raft at a distance:

S1 = (v1 + v0) * t3.

During the time after meeting with the motor boat, the raft covered a distance equal to:

S2 = v0 * t2 + v0 * t3.

It’s obvious that:

S1 = S + S2;

(v1 + v0) * t3 = (v1 – v0) * t1 + v0 * t2 + v0 * t3;

v1 * t3 + v0 * t3 = v1 * t1 – v0 * t1 + v0 * t2 + v0 * t3;

v1 * t3 – v1 * t1 = v0 * t2 + v0 * t3 + v0 * t1 – v0 * t3;

v1 * (t3 – t1) = v0 * (t2 + t3 + t1 – t3);

v1 * (t3 – t1) = v0 * (t2 + t1);

v1 = v0 * (t2 + t1) / (t3 – t1) = 0.5 * (2100 + 1200) / (2400 – 2100) =

= 0.5 * 3300/300 = 1650/300 = 5.5 m / s.

Answer: the speed of the motor boat relative to the water is 5.5 m / s.