# Question about total vector displacement, BUGGING ME OUT

If I were to drive south at 20.0 m/s for 3 min, then turn west and travel at 25 m/s for 2 min, and finally travel northwest at 30.0 m/s for 1.00 min. what would my total vector displacement be?

I start out by converting minutes to seconds, so I have position vectors; 1800m to south, 3600m west, and 1800m NorthWest. Then using the respective vector units, I add them up, and square each component, and take the square root. I keep getting a different answer, please help.

Displacement Vector = Sqrt ( (3600 +1800cos45 + 0)i ^2 + ((1800sin45 +3000)^2 j))

Am i on the wrong track here, this is REALLY bugging me out!!!

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Hootenanny
Staff Emeritus
Gold Member
Welcome to PF,
Displacement Vector = Sqrt ( (3600 +1800cos45 + 0)i ^2 + ((1800sin45 +3000)^2 j))
You need to sort your unit vectors out. Here you have north & south and east & west as the same direction.

Usually, i would be directed from west to east and j would be directed from south to north.

How would I go about finding the average speed? I figured out that average speed is the magnitude of the average velocity vector = Delta (R)/Delta(T), so How does that apply to this problem?

bummp, anyone?

How would I go about finding the average speed? I figured out that average speed is the magnitude of the average velocity vector = Delta (R)/Delta(T), so How does that apply to this problem?
Average speed is total distance/time. Not always the same as average velocity.