# 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?

By BYJU'S Exam Prep

Updated on: September 25th, 2023

(a) 22km/hr, at an angle tan-1 (0.343) North of East

(b) 22km/hr, at an angle tan-1 (0.433) East of North

(c) 20km/hr, at an angle tan-1 (0.433) East of North

(d) 20km/hr, at an angle tan-1 (0.343) North of East

The resultant velocity of the boat is 22km/hr, at an angle tan-1 (0.433) East of North. The vectors vb and vc in the figure are the motorboat’s velocity and the water current, respectively, in the directions suggested by the issue.

The resulting R is derived by adding using the parallelogram approach in the direction depicted in the image.

### 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.

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