A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 600 east of south. The resultant velocity of the boat is?

Find the Resultant Velocity of the Boat

Consequently, the magnitude of the outcome is provided by:

R = √vb2 + vc2 + 2vbvc cos θ where θ = 1200

Put the values in the above equation we get

R = √252 + 102 + 2 x 25 x 10 cos (1200)

In simplification we get the:

R = 21.8 ≈ 22 km/hr

Now, for the direction we have:

tan ϕ = vc sin(1200)/ vb + vc (1200) = 10 sin (1200)/ 25 + 10 cos (1200)

In simplification we get the:

tan ϕ = 0.433

ϕ = tan-1 (0.433) East of North

Summary:

A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 600 east of south. The resultant velocity of the boat is?

A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction of 600 east of south. The resultant velocity of the boat is 22km/hr, at an angle tan-1 (0.433) East of North.