A plane is flying southwest at 155 mi/h Suddenly there is a wind from the west at 45.0 mi/h. what is the [lane new velocity with respect to the ground in standard position?

This is a problem in vector addition.

Resolve the two velocity vectors into rectangular components. Add the respective components. Then reconvert that sum back to polar coordinates.

If this needs more explanation, please post back.

We can solve it this way too, without vectors.

Velocity = SQRT[(X-Component)^2 +(Y-Component)^2]

Flying South-West. Hence making an angle of 45 degrees with West as well as with South.

Original Velocity component toword South: 155 Cos 45
Original Velocity component toword West : 155 Sin 45

NEW

Velocity component toword South: 155 Cos 45
Velocity component toword West: 155 Sin 45 - 45

New Velocity = SQRT [(155 Cos 45)^2 + (155 Sin 45 - 45)^2]

= 127 miles per hour

Jiten,

This ignores that velocity is by definition a vector.

Your result is not a requested velocity, but in in fact a scalar, speed.
Scalar - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Scalar)

Also, my intention was to allow the OP to do the math for themselves.

The final answer does still require a direction.

PolluxCastor:

Technically you are correct, it is a vector.

But my experience is that at junior levels all they want is the speed when they talk of velocity.

We should let the querist decide the matter!

Of course if direction is required, it can be calculated from the 2 components of velocity which I stated.

If the querist needs further help he can always seek for further elucidation.