Question

One Ship Is Sailing South With A Velocity Of 15√2 Km/H And Another With 15 Km/Hr Towards South East. The Relative Velocity Of Second Ship With Respect To First Ship

Answer 1

Let A and B are two ships. ship A is sailing south with a velocity of 15√2 km/h. so, velocity of ship A in vector form, $v_A$ = -15√2 j km/h

again, another ship B is sailing towards south east with a velocity of 15km/h.

so, velocity of ship B in vector form, $v_B$ = 15(cos45° i - sin45° j)

= 15/√2 i - 15/√2 j

now, relative velocity of ship B with respect to ship A, $v_{BA}=v_B-v_A$

= 15/√2 i - 15/√2 j - (-15√2 j )

= 15/√2 i - (15/√2 - 15√2)j

= 15/√2 i - (15 - 30)/√2 j

= 15/√2 i + 15/√2 j

magnitude of $v_{AB}$ = √{(15/√2)² + (15/√2)²}

= √{225/2 + 225/2} = 15 km/h.

angle made by relative velocity is 45° with positive direction of x - axis or direction of relative velocity is N-E.

Answer 2

Answer:

The relative velocity of second ship with respect to first ship is 15 km/hr in N-E.

Explanation:

Given that,

Velocity of ship A $v_{A}= 15\sqrt{2}\ km/h$

Velocity of ship B $v_{B}=15\ km/h$

We need to calculate the relative velocity of second ship with respect to first ship

Using formula of relative velocity

$v_{AB}=\sqrt{v_{A}^2+v_{B}^2-2v_{A}v_{B}\cos\theta}$

Put the value into the formula

$v_{AB}=\sqrt{(15\sqrt{2})^2+(15)^2-2\times15\sqrt{2}\times15\cos45}$

$v_{AB}=15\ km/h$

The direction is north - east.

Hence, The relative velocity of second ship with respect to first ship is 15 km/hr in N-E.