Vectors: How to Calculate Resultant Velocity and Direction in a Crosswind?

[max1995]

Question
An object is traveling eastward at a speed of 500 ms-1 flies into a 90ms-1 crosswind blowing southward.
a) calculate the resultant velocity of the plane relative to the object as it flies through the crosswind.

b) A person is steering the object, calculate the direction the person would have to steer in order for the resultant velocity off the plane to remain eastward.

My attempt

a) Magnitude of R=squareroot(902+500²)
= 508ms-1

Direction= tan-1(90/500) = 10.2
90-10.2= 79.8

Resultant= 508ms-1 at 79.8 degrees to the ground

b) don't know what to do, would you please be able to help me on this one? I don't really understand what it wants me to do

Thanks for any help you can give

 

[Bystander]

Plane? Taxiing? Parked? Sitting in a hangar? Are the "plane" and the "object" one and the same?

 

[max1995]

Plane? Taxiing? Parked? Sitting in a hangar? Are the "plane" and the "object" one and the same?

Sorry missed the bit at the bottom its a plane

 

[BvU]

Make a drawing for case a)
Make a drawing for the case b)

 

[max1995]

Bump still don't know how to do part b

 

[Bystander]

For b) you want the plane/object moving at 500m/s to travel true east while the wind is blowing S at 90 m/s. What velocity N is necessary to offset the wind? What velocity E combined with that gives you a 500 m/s resultant air speed?

 

[max1995]

For b) you want the plane/object moving at 500m/s to travel true east while the wind is blowing S at 90 m/s. What velocity N is necessary to offset the wind? What velocity E combined with that gives you a 500 m/s resultant air speed?

Thank you for the help
so I want to find a velocity (with angle north) that will give me 500km h-1 east?

if what I said above is right, I would

500= sqaureroot(90² +X²) then doing some rearranging and calculating

X= 491.83 km h-1

then to work out the direction would do

500= 90 + 491.83cos(y)
y= cos^-1(410/491.83)
y= 33.53 degrees to the eastward

therefore the direction and speed that the plane would need to travel is 491.83km h-1 at an angle of 33.53 degrees to the eastwood?

is that correct

thank you again for the help

 

[Bystander]

(snip)

500= sqaureroot(90² +X²) then doing some rearranging and calculating

X= 491.83 km h-1

My calculator just cratered --- I'll take your word --- it looks close.

then to work out the direction would do

500= 90 + 491.83cos(y)

No. BvU suggested you make drawings. Which part of the right triangle is 500 for this part of the problem? Keep in mind that part b) may be different than part a).

 

[max1995]

My calculator just cratered --- I'll take your word --- it looks close.
No. BvU suggested you make drawings. Which part of the right triangle is 500 for this part of the problem? Keep in mind that part b) may be different than part a).

so the 500 is the resultant? making 491.83=500sin(y)?? and this makes y to the north?

is that correct

 

[Bystander]

so the 500 is the resultant? making 491.83=500sin(y)?? so then y is the angle to the north?

iis that correct

Yes. And yes, or cos(y), whichever you prefer. You measure angles from a reference direction in either clockwise (navigation), or counterclockwise (right handed coordinate systems in mathematics) direction; what reference direction are you using, and are you measuring cw or ccw?