Question

$\huge{\mathfrak{Hello\;Frnds}}$

Here's ur question!!

Ship A is moving to east with a speed 10km/h. Ship B is moving relative to A 30° North-east, such that what should be the Velocity of B in order to keep it always due to North...

Content quality ✔️
No Spamming ❌
_____________________❤️
​

Answer 1

Answer:

20 km / hr.

Explanation:

Let u is the velocity of ship A and v is velocity of ship B.

We have already given velocity of ship A i.e. u = 10 km / hr.

Now , Slitting the components we get :

v cos 60 for ship A and v sin 30 for ship B.

In order to move ship B always North-east w.r.t. A  drift should be zero.

Horizontal component is equal to velocity of ship A

i.e.  v cos 60 = u    [ u = 10 km / hr ]

v = 10 / 1 / 2 km / hr

v = 20 km / hr.

Hence the velocity should be 20 km / hr of ship B in order to keep it always due to North.

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• The velocity of Ship "A" is 10 Km/hr.
• Angle made is 30°.

$\huge{\bold{\underline{Explanation:-}}}$

#refer the attachment for figure,

Let the velocity of Ship "A" be $\large{\vec{v_1}}$.

and, The velocity of Ship "B" be $\large{ \vec{v_2}}$

As we know,

it is a right angled triangle,

As the it is right angled at "C".

${ \angle{C} = 90}$

Applying trigonometric functions,

Applying Sine angle formula,

$\large{\boxed{ \tt \sin \theta = \dfrac{\vec{v_1}}{\vec{v_2}}}}$

Substituting the values,

${ \theta = 30}$

Now,

$\large{\sin 30 = \dfrac{10}{ \vec{v_2}}}$

As sin 30° is ½.

It becomes,

$\large{ \dfrac{1}{2} = \dfrac{10}{\vec{v_2}}}$

Cross multiplying,

$\large{ \vec{v_2} = 2 \times 10}$

$\huge{\boxed{\boxed{ \tt {\vec{v_2}} = 20 \: Km/hr}}}$

So, the Velocity of B in order to keep it always due to North is 20 Km/hr.