Question

A swimmer can travel 2.30 m/s in still water. The swimmer heads directly west across a steam whose current is 1.20 m/s n. What is his velocity relative to the ground

Answer 1

Answer:

velocity relative to the ground = 2.54 m/s

Explanation:

given data

still water Vx = 2.30 m/s

across a steam Vy = 1.20 m/s

solution

we get here first velocity relative to the ground is express as

velocity = $\sqrt{Vx^2+Vy^2}$      ...............1

put here value and we get

velocity  = $\sqrt{2.3^2+1.20^2}$

velocity = 2.54 m/s