# Ground speed of a plane - vectors

## Homework Statement

A plane is heading due east and climbing at he rate of 80kph. if its speed is 480kph and there's a wind blowing 100kph to the northeast, wha is the groundspeed of the plane

## The Attempt at a Solution

w=100cos45+100sin45= 70i+70j
480= vxi+ 70j+80k
vxi= 468
Groundspeed= 468i+70i+70j=538i + 70j = 543kph

or

480= vxi+70j
vxi= 475
Groundspeed = 475i +70i+70j = 545i + 70j = 549kph

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HallsofIvy
Homework Helper

## Homework Statement

A plane is heading due east and climbing at he rate of 80kph. if its speed is 480kph and there's a wind blowing 100kph to the northeast, wha is the groundspeed of the plane

## The Attempt at a Solution

w=100cos45+100sin45= 70i+70j
480= vxi+ 70j+80kk
How do you get the "70j" here?

vxi= 468
Groundspeed= 468i+70i+70j=538i + 70j = 543kph

or

480= vxi+70j
vxi= 475
Groundspeed = 475i +70i+70j = 545i + 70j = 549kph
Again, where did you get the "70 j" as part of the airplane's velocity? If the airplane is going "due east" shouldn't it be 0j?

yea i thought as much but isn't the y-component of the windspeed part of the airspeed? Can you show me how to solve this problem?

HallsofIvy
Homework Helper

No, the "airspeed" is the speed through the air and is separate from the wind speed. And you surely can't have thought you should include the j (north-south) component but not the i (north-south) component?

The velocity relative to the air is vx i+ 80 k and the airspeed is 480 so $vx^2+ 80^2= 480^2$. Once you have found that, the velocity relative to the ground is (vx+ 70)i+ 70j (the k component is not relevant to moving relative to the ground).

ok thanks, i got 547.8kph. the answer key is 548.6, i guess is still correct...

Mark44
Mentor

You lost some precision by the very rough rounding you did early on.
100 $\sqrt{2}$/2 is closer to 71 than it is to 70.

You will always get more precise results if you refrain from rounding until your final result.

Thanks, it was a complete oversight...duly noted