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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.