Ship A is travelling with a velocity of 5km h−1 due east. The second ship is heading 30∘ east of north. The speed of second ship if it is to remain always due north with respect to the first ship in km h−1 is
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Solution
For B always to be north of A, the velocity components of both along east should be same. v2cos60∘=v1 ⇒v2=v1cos60∘ ⇒v2=2×v1 ⇒v2=10km/h