A ship moves along the equator to the east with a speed 30 km-h- Southeastern wind blows 60 - to the east with 15 km h -1- Find the wind velocity relative to the ship-

The correct option is A 41 km h ^ 1 ,tan ^ 1 √(3) 5 North of WestRelative velocity is v_w v_s ( 15cos60î + 15sin60 #160; ĵ) 30î so ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Relative Velocity

A ship moves ...

Question

A ship moves along the equator to the east with a speed $30 k m / h$ . Southeastern wind blows $60^{\circ}$ to the east with $15 k m h^{- 1}$ . Find the wind velocity relative to the ship.

41 k m h^{- 1}, {tan}^{- 1} \frac{\sqrt{3}}{5}

North of West

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

23.7 k m h^{- 1}, {tan}^{- 1} \frac{1}{3}

North of West

No worries! We‘ve got your back. Try BYJU‘S free classes today!

37.5 k m h^{- 1}, {tan}^{- 1} \frac{1}{5}

North of West

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $41 k m h^{- 1}, {tan}^{- 1} \frac{\sqrt{3}}{5}$ North of West
Relative velocity is $v_{w} - v_{s}$
$(- 15 c o s 60^i + 15 s i n 60^j) - 30^i$

so $| V | = \sqrt{{37.5}^{2} + {12.99}^{2}} = 39.98 m / s e c$

and we know that $t a n θ = \frac{y}{x}$
$t a n θ = \frac{12.99}{37.5} = \frac{\sqrt{3}}{5}$ in North West

Suggest Corrections

Similar questions

A political party has to start its procession in an area where wind is blowing at a speed of 41.4 km/h and party flags on the vehicles are fluttering along north-east direction. If the procession starts with a speed of 40 km/h towards north, find the direction of flags on the vehicles.

Q. Ship

A

is travelling with a velocity of

5 {km h}^{- 1}

due east. The second ship is heading

30^{\circ}

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}

Q. A ship moves along the equator to the east with velocity

v_{0} = 30 k m / h

. The southeastern wind blows at an angle

φ = 60^{\circ}

to the equator with velocity

v = 15 k m / h

. The angle

φ^{'}

between the equator and the wind direction in the reference frame fixed to the ship is (in degrees)

Q. The wind appears to blow from north to a man moving in the north-east direction. When he doubles his velocity, the wind appears to move in the direction cot

^{- 1}

(2) east of north. If the actual magnitude of velocity of the wind is

V = \frac{1}{\sqrt{x}} \times

magnitude of velocity of man and direction along east. Find

x

A bird is flying due east with a velocity of 4 m/s. The wind starts to blow with a velocity of 3 m/s due north. What is the magnitude of relative velocity of bird w.r.t. wind? Find out the angle it makes with the x-axis?

Join BYJU'S Learning Program

A ship moves along the equator to the east with a speed 30 km/h. Southeastern wind blows 60∘ to the east with 15 kmh−1. Find the wind velocity relative to the ship.

The correct option is A 41 kmh−1,tan−1√35 North of WestRelative velocity is vw−vs(−15cos60^i+15sin60^j)−30^iso |V|=√37.52+12.992=39.98m/secand we know that tanθ=yxtanθ=12.9937.5=√35 in North West

A ship moves along the equator to the east with a speed $30 k m / h$ . Southeastern wind blows $60^{\circ}$ to the east with $15 k m h^{- 1}$ . Find the wind velocity relative to the ship.