Question

A cruise ship sails due south at 2.00 m/s while a coast guard patrol boat heads 19.0° north of east at 5.60 m/s. What are the x-component and y-component of the velocity of the cruise ship relative to the patrol boat? (Assume that the +x-axis is east and the +y-axis is north. Enter your answers in m/s.) HINT

Answer 1

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Step-by-step explanation:

Given that,

Velocity of ship = 2.00 m/s due south

Velocity of boat = 5.60 m/s due north

Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

`v_{s_(x)}=0`

`v_{s_(y)}=2.0\ m/s`

The velocity of the boat in term x and y coordinate

For x component,

`v_{b_(x)}=v_(b)\cos\theta`

Put the value into the formula

`v_{b_(x)}=5.60\cos19`

`v_{b_(x)}=5.29\ m/s`

For y component,

`v_{b_(y)}=v_(b)\sin\theta`

Put the value into the formula

`v_{b_(y)}=5.60\sin19`

`v_{b_(y)}=1.82\ m/s`

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

`v_{sb_(x)}=v_{s_(x)}-v_{b_(x)}`

Put the value into the formula

`v_{sb_(x)=0-5.29`

`v_(sb)_(x)=-5.29\ m/s`

For y component,

`v_{sb_(y)}=v_{s_(y)}-v_{b_(y)}`

Put the value into the formula

`v_{sb_(x)=2.-1.82`

`v_(sb)_(x)=0.18\ m/s`

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.