Question

A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00º west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.)

Answer 1

Answer:

Velocity of Gulf Stream= $34.5^\circ$ west of north at a speed of $2.03\ \rm m/s$

Explanation:

Given

• speed of ship relative to Gulf stream=$25^\circ$  west of north at a speed of 4 m/s

• speed of ship relative to earth=$5^\circ$  west of north at a speed of 4 .8 m/s

Let $V_S$ be the velocity of the ship relative to the earth and $V_{sg}$ be the velocity of the ship with respect to the Gulf stream

Let the west direction be  negative x axis and north direction be positive x axis

Now

$V_{sg}=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\V_s-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\-4.8\sin5^\circ \vec i+4.8\cos5^\circ \vec j-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\Vg=-1.68\vec\ i+1.156\vec\ j\ \rm m/s$

magnitude of the velocity is given by

$V_g=\sqrt{(1.68^2+1.156^2)}\\V_g=2.03\ \rm m/s$

Let the velocity of gulf stream makes an angle $\theta$ with the positive y axis we have

$\tan\theta=\dfrac{1.156}{1.68}\\\theta=34.5^\circ$

Velocity of Gulf Stream $34.5^\circ$ west of north at a speed of $2.03\ \rm m/s$

Answer 2

The velocity of the gulf stream is $77.76^\circ$ west of north at a speed of 1.735 m/s.

How do you calculate the velocity?

Given that A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00º west of north.

Let's consider that Vs is the velocity of Ship and Vg is the velocity of Gulf. Vsg is the velocity of the ship with respect to the gulf.

The ship is moving in the west direction, so we will consider the direction as per the XY coordinate. So west direction will fall under negative X-Axis. Then,

$V_s = -4sin 5^\circ \vec{i} + 4cos 5^\circ \vec{j}$

$V_{sg} = -4.8 sin25^\circ \vec{i} + 4.8cos 25^\circ \vec{j}$

$V_{sg} = V_s -V_g$

Substituting the values, we get Vg.

$-4.8sin25^\circ \vec{i} + 4.8cos 25^\circ \vec{j} = -4sin 5^\circ \vec{i} + 4cos 5^\circ \vec{j} - V_g$

$V_g = -1,696 \vec{i} + 0.368 \vec {j}$

The magnitude can be calculated as given below.

$V_g = \sqrt{1.696^2 + 0.368^2}$

$V_g = 1.735 \;\rm m/s$

Let's consider that the velocity of the gulf stream makes an angle $\theta$ with the positive y-axis, then,

$tan \theta = \dfrac {1.696}{0.368}$

$tan\theta = 4.61$

$\theta = 77.76^\circ$

Hence we can conclude that the velocity of the gulf stream is $77.76^\circ$ west of north at a speed of 1.735 m/s.

To know more about velocity, follow the link given below.

https://brainly.com/question/862972.