Question

Ship a is travelling with a velocity of 5 km/h due east. A second ship is heading 30° east of north. What should be the speed of second ship if it is to remain always due north with respect to the final ship

Answer 1

hence, speed of second ship is 10 km/h.

Let speed of second ship is x km/h

velocity of first ship , v1 = 5 i km/h

velocity of second ship , v2 = x(sin30° i + cos30° j)

= x( 1/2 i + √3/2 j) km/h

as velocity of second ship with respect to first ship is always due to north.

i.e., v21 does has only y-component.

here, v21 = v2 - v1

= x(1/2 i + √3/2 j) - 5i

= (x/2- 5) i + √3x/2 j

as v21 has only y - component. then x - component of (x/2- 5) i + √3x/2 j should be zero.

i.e., x/2 - 5 = 0 ⇒x = 10 km/h

hence, speed of second ship is 10 km/h

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Answer 2

$\huge\bold\purple{Answer:-}$

Let speed of second ship is x km/h

velocity of first ship , v1 = 5 i km/h

velocity of second ship , v2 = x(sin30° i + cos30° j)

= x( 1/2 i + √3/2 j) km/h

as velocity of second ship with respect to first ship is always due to north.

i.e., v21 does has only y-component.

here, v21 = v2 - v1

= x(1/2 i + √3/2 j) - 5i

= (x/2- 5) i + √3x/2 j

as v21 has only y - component. then x - component of (x/2- 5) i + √3x/2 j should be zero.

i.e., x/2 - 5 = 0 ⇒x = 10 km/h

hence, speed of second ship is 10 km/h