Question

One ship is sailing south with a velocity of
15root2 km/hr and another South-East at the rate of
15 kmph. The relative velocity of second ship with
respect to the first ship is
(1) 10 km/hr S-W
(2) 15 kmhr S-w
(3) 10 km/hr N-E
(4) 15 km/hr N-E
Explanation.​

Answer 1

Here is your answer

I hope it helps

Answer 2

Answer:

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

(4) is correct option.

Explanation:

Given that,

Speed of first ship in south $v= 15\sqrt{2}\ km/hr$

Speed of another ship in south - east $v'=15\ km/hr$

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

Using formula of relative velocity

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

Here, $\theta = 45^{\circ}$

Put the angle in to the formula

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

$\vec{v_{AB}}=15\ km/hr$

The direction of relative velocity is north - east.

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