help needed (motion)

bluestreak

Question..
a plane heading due west has an airspeed of 525km/h, the wind is blowing at a rate of 100km/h from the south west...

calculate ground speed ???

calculate the angle the plane has been blown off course????

Kohl

bluestreak wrote: »

Question..
a plane heading due west has an airspeed of 525km/h, the wind is blowing at a rate of 100km/h from the south west...

calculate ground speed ???

calculate the angle the plane has been blown off course????

I think this is a vectors question. I need to dig out my notes on this one. We have v1 = - 525 i for the plane and we need the vector for the wind which is

v2 = 100 cos (45) i + 100 sin (45) j = 50 sqrt(2) i + 50 sqrt(2) j.

I think you add these then to get resultant

v = (-525 + 50 sqrt(2) ) i + 50 sqrt(2) j

To get your angle dividethe value of j by the value of i in the resultant and take the arctan to get the angle. I get theta = 8.8 degrees.

I don't know how to do the groundspeed part

bluestreak

thanks for the help, hopefully i will get there in the end

Morbert

Kohl wrote: »

I think this is a vectors question. I need to dig out my notes on this one. We have v1 = - 525 i for the plane and we need the vector for the wind which is

v2 = 100 cos (45) i + 100 sin (45) j = 50 sqrt(2) i + 50 sqrt(2) j.

I think you add these then to get resultant

v = (-525 + 50 sqrt(2) ) i + 50 sqrt(2) j

To get your angle dividethe value of j by the value of i in the resultant and take the arctan to get the angle. I get theta = 8.8 degrees.

I don't know how to do the groundspeed part

The groundspeed will be the magnitude of the vector v.

bluestreak

think ive just done it,
YOU draw both vectors head to trail, and the resultant is your vector.
or by using Pythagoras therm ,525^2+100^2=r^2 which is 285625 and the square root of that equals 534.43km/and to get the angle you get the inverse tan of 100/525 which equals 10.78 degrees

thanks for your help

sponsoredwalk

bluestreak wrote: »

think ive just done it,
YOU draw both vectors head to trail, and the resultant is your vector.
or by using Pythagoras therm ,525^2+100^2=r^2 which is 285625 and the square root of that equals 534.43km/and to get the angle you get the inverse tan of 100/525 which equals 10.78 degrees

thanks for your help

The Pythagorean theorem only applies if the vectors are perpendicular.
If the plane was heading West & the wind was blowing North then you'd
have the correct answer but because the wind is blowing from the South
West towards the North East the vectors aren't perpendicular.

Since the wind is blowing from the South West to the North East it will
blow at a certain speed, which can be represented by a vector [latex]v_w[/latex].
Just draw that vector at the origin of your coordinate system, i.e.
starting at O and ending at some point P. Hopefully you can figure out
the angle that this vector is making with respect to the x-axis.
Now, since there is a plane heading West it will also have a velocity that
is representable as a vector so also draw that vector starting from O &
ending at some point Q.

Since vectors can be moved anywhere you want so long as you don't
change the direction or the length (magnitude) you can, by the properties
of vector addition, move one of the vectors tails to the head of the
other vector and, by drawing an arrow from point O to the tip of the
second arrow form a new vector, which is the sum of the other two
& this represents the final velocity. What you're doing is adding to the
velocity of the plane a new velocity in a new direction created by the
wind.

Once you've done this you can use the Law of Cosines to find the
information you require or you can decompose both vectors into
their component form (i.e. [latex]v_x \ \ihat{i} \ + \ v_y \ihat{j}[/latex]) and add them
to get the final velocity.

The "ground speed" really just means "speed" which is the magnitude
of the velocity vector.

im invisible

sponsoredwalk wrote: »

The "ground speed" really just means "speed" which is the magnitude of the velocity vector.

does it not depend how high the plane is flying :cool:

mawk

im invisible wrote: »

does it not depend how high the plane is flying :cool:

only if the plane is pitching vertically up or down. then the groundspeed is tending to zero and the airspeed is however fast its actually going.
I assume the original question assumes the plane is flying dead level.







:cool: