Question

A boat travels at 15 m/s in a direction 45° east of north. The boat then turns and travels at 18 m/s in a direction 5° north of east.

What is the magnitude of the boat’s resultant velocity? Round your answer to the nearest whole number.

m/s
What is the direction of the boat’s resultant velocity? Round your answer to the nearest whole degree.

° north of east

Answer 1

The magnitude of the boat's resultant velocity is 15.8 m/s, and the direction is 130 degrees north of east.

Step-by-step explanation:

To find the magnitude of the boat's resultant velocity, we can use the law of cosines. The resultant velocity is the vector sum of the velocities of the boat in the two given directions. Using the given information, we can calculate the magnitude of the resultant velocity:

Magnitude = sqrt((15^2 + 18^2) - 2(15)(18)cos(45-5)) = 15.8 m/s (rounded to the nearest whole number).

To find the direction of the boat's resultant velocity, we can use trigonometry. The direction is the angle between the resultant velocity vector and due east. Using the given information, we can calculate the direction of the resultant velocity:

Direction = 45 + (90 - 5) = 130 degrees north of east (rounded to the nearest whole degree).

Answer 2

1) Magnitude

Let's take north as positive y-direction and east as positive x-direction. Then we have to resolve both velocities into their respective components:

`v_(1x) = (15 m/s) sin 45^(\circ)=10.6 m/s`

`v_(1y) = (15 m/s) cos 45^(\circ)=10.6 m/s`

`v_(2x) = (18 m/s) cos 5^(\circ)=17.9 m/s`

`v_(2y) = (18 m/s) sin 5^(\circ)=1.6 m/s`

So, the components of the resultant velocity are

`v_x = v_(1x)+v_(2x)=10.6 m/s+17.9 m/s=28.5 m/s` east

`v_y=v_(1y)+v_(2y)=10.6 m/s+1.6 m/s=12.2 m/s` north

So, the magnitude of the resultant velocity is

`v=√(v_x^2+v_y^2)=√((28.5)^2+(12.2)^2)=31.0 m/s`

2) Direction

the direction of the boat's velocity is

`\theta= arctan((v_y)/(v_x))=arctan((12.2)/(28.5))=arctan(0.428)=23.2^(\circ)` north of east