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

When resolving the car's velocity vector into components, we need to consider the directions of travel. If the car is traveling east, the velocity vector can be resolved into 2 components: one in the positive x-direction and one in the y-direction, which is 0. If the car is traveling south, the velocity vector can be resolved into 2 components: one in the positive y-direction and one in the x-direction, which is 0. If the car is traveling southeast, the velocity vector can be resolved into 2 components: one in the positive x-direction and one in the negative y-direction. If the car is traveling northwest, the velocity vector can be resolved into 2 components: one in the negative x-direction and one in the positive y-direction.

Step-by-step explanation:

When resolving the car's velocity vector into components, we need to consider the directions of travel.

If the car is traveling east, the velocity vector can be resolved into 2 components: one in the positive x-direction and one in the y-direction, which is 0.

If the car is traveling south, the velocity vector can be resolved into 2 components: one in the positive y-direction and one in the x-direction, which is 0.

If the car is traveling southeast, the velocity vector can be resolved into 2 components: one in the positive x-direction and one in the negative y-direction.

If the car is traveling northwest, the velocity vector can be resolved into 2 components: one in the negative x-direction and one in the positive y-direction.

Answer 2

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`