A ship ,4 sailing due east with a velocity of 10 km/h happens to appear sailing due north with a velocity of 5 km/h , to a person, sitting in a moving ship B. Determine the velocity (absolute) of ship B.
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a
55km/h,tan−1(1/2)S of E
b
55km/h,tan−1(1/2)E of S
c
45km/h,tan−1(1/2)S of E
d
45km/h,tan−1(1/2)E of S
answer is A.
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Detailed Solution
Here we are given velocity of ‘A’, v→A=10i^Velocity of 'A' w.r.t ‘B’, v→A/B=5j^Now v→A/B=v→A−v→B5j^=10i^−v→B⇒v→B=10i^−5j^Hence velocity of B,vB=102+52=55km/htanθ=510=12θ=tan−112S of E