Question

Suppose a car is traveling at 65 km/hr, and that the positive y-axis is north and the positive x-axis is east. Resolve the car's velocity vector (in 2-space) into components if the car is traveling in each of the following directions.
East: .
South: .
Southeast: .
Northwest: .

Answer 1

Answer:

East:

$v_{x}=65km/h\\v_{y}=0km/h$

South:

$v_{x}=0km/h\\v_{y}=-65km/h$

Southeast:

$v_{x}=46km/h\\v_{y}=-46km/h$

Northwest:

$v_{x}=-46km/h\\v_{y}=46km/h$

Step-by-step explanation:

The vector components can be express as follows:

$v_{x}=v*cos(\theta)\\v_{y}=v*sin(\theta)$

Where $v$ is the velocity of the car, $v_{x}$ is the x-component of the velocity, $v_{y}$ is the y-component of the velocity and, $\theta$ is the angle between the velocity vector the positive x-axis (angle increases anti clockwise).

Before computing the components of velocity we need to remember that:

• East lays over the positive x-axis.

• South lays over the negative y-axis.

• Southeast is in between South and East.

• Northwest is in between North and West.

For East we have that $\theta=0$ so,

$v_{x}=65*cos(0)\\v_{y}=65*sin(0)$

$v_{x}=65km/h\\v_{y}=0km/h$

for South we have that  $\theta=270$ so,

$v_{x}=65*cos(270)\\v_{y}=65*sin(270)$

$v_{x}=0km/h\\v_{y}=-65km/h$

for Southeast we have that $\theta=315$ so,

$v_{x}=65*cos(315)\\v_{y}=65*sin(315)$

$v_{x}=46km/h\\v_{y}=-46km/h$

and finaly, for Northwest we have that $\theta=135$ so,

$v_{x}=65*cos(135)\\v_{y}=65*sin(135)$

$v_{x}=-46km/h\\v_{y}=46km/h$