# The boat is crossing the river 800m wide perpendicular to the current, the speed of which is 6 km / h.

**The boat is crossing the river 800m wide perpendicular to the current, the speed of which is 6 km / h. The speed of the boat relative to the water is 10 km / h. What is the speed of the boat relative to the coast? How long will the crossing take? How far will the boat blow downstream?**

S = 800 m = 0.8 km.

VT = 6 km / h.

Vк = 10 km / h.

Vkb -?

t -?

d -?

According to the law of relativity of movement, the speed of the boat relative to the coast Vкб will be the vector sum of the boat’s own speed Vк and the speed of the current Vт: Vкб = Vк + Vт.

Since the boat’s own speed Vk is directed perpendicular to the current velocity Vt, then according to the Pythagorean theorem: Vkb = √ (Vk ^ 2 + Vt ^ 2).

Vkb = √ ((10 km / h) ^ 2 + (6 km / h) ^ 2) = 11.7 km / h.

The remelting time t is expressed by the formula: t = S / Vк.

t = 0.8 km / 10 km / h = 0.08 h = 4.8 min = 288 s.

The boat drift distance d is expressed by the formula: d = Vt * t.

d = 6 km / h * 0.08 h = 0.48 km = 480 m.

Answer: Vkb = 11.7 km / h, t = 0.08 h = 4.8 min = 288 s, d = 480 m.