The swimmer, whose speed relative to the water is 1.4 m / s, crosses the river 120 m wide, moving perpendicular

The swimmer, whose speed relative to the water is 1.4 m / s, crosses the river 120 m wide, moving perpendicular to the current. The current speed is 0.9 m / s. What is the swimmer’s speed and movement relative to the shore?

Vp = 1.4 m / s.

d = 120 m.

Vt = 0.9 m / s.

S -?

Vpb -?

Let’s find the time t, during which the swimmer swims across the river according to the formula: t = d / Vп.

t = 120 m / 1.4 m / s = 86 s.

During the same time t, it will be swept away by the current along the coast to a distance b = Vt * t.

b = 0.9 m / s * 86 s = 77.4 m.

The swimmer’s movement S is the vector that connects his starting and ending position. This will be the hypotenuse of a right-angled triangle with legs d and b. By the Pythagorean theorem, S = √ (d ^ 2 + b ^ 2).

S = √ ((120m) ^ 2 + (77.4m) ^ 2) = 142.8m.

The swimmer’s speed relative to the shore is expressed by the formula: Vpb = √ (Vp ^ 2 + Vt ^ 2).

Vpb = √ ((1.4 m / s) ^ 2 + (0.9 m / s) ^ 2) = 1.7 m / s.

Answer: S = 142.8 m, Vpb = 1.7 m / s.