How Do You Calculate Total Vector Displacement and Average Velocity?

[-EquinoX-]

Homework Statement

A motorist drives south at 20.0 m/2 for 3.00 min, then turns west and travels at 25.0 m/s for 2 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00 min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity.

Homework Equations

The Attempt at a Solution

Well I tried to convert all that to distance, so multiplying velocity and time. What I don't understand is the north east part. If I were to represent the first part. it would be 3600 m to the south, so -3600j and the second one would be -3000i to and what about the north east part?

 

[LowlyPion]

Homework Statement

A motorist drives south at 20.0 m/2 for 3.00 min, then turns west and travels at 25.0 m/s for 2 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00 min trip, find (a) the total vector displacement, (b) the average speed, and (c) the average velocity.

Homework Equations

The Attempt at a Solution

Well I tried to convert all that to distance, so multiplying velocity and time. What I don't understand is the north east part. If I were to represent the first part. it would be 3600 m to the south, so -3600j and the second one would be -3000i to and what about the north east part?

The problem says North West.

The length of the vector will be 1800 but the component vectors would be multiplied by (1/2)*sqrt(2).

 

[-EquinoX-]

where did you get (1/2)*sqrt(2) from?

 

[LowlyPion]

where did you get (1/2)*sqrt(2) from?

That's the value of sine and cosine of a 45 degree angle. A right triangle of unit sides has a hypotenuse of sqrt(2). So finding the sine and cosine is simply 1/sqrt(2) or it can be written as (1/2)*sqrt(2).

You are going Northwest so the sine and cosine components of the displacement vector at that angle can be given as that.

 

[-EquinoX-]

so if it says north west then you're assuming the angle is 45 degress with the x axis