Need help with physics question, please!

Hi Loza, I found a similar question and its answer, I hope this helps

Okay, I finished answering this with 170 as the velocity and a velocity of the air with respect to the ground of 32 m/s.

The velocity of the plane with respect to the air v = 170 i m/s
The velocity of the air with respect to the ground v ' = 32 cos 30 x (-i) + 32 sin 30 x ( j )

= -27.712 i + 16 j

1)The velocity of the plane with respect to the ground v " = v - v '

=197.712 i -16 j

The speed of the plane with respect to the ground v " = SQRT[ 197.712 2 + (-16) 2 ]

= 198.358 m / s

2)The heading of the plane with respect to the ground θ = tan - 1( 16 / 197.712 )

= 4.626 degrees south of east

3)Time t = 1 h

= 3600 s

Distance travel by plane in east direction with in 1 hour time = Speed in east x time

= 197.712 m / s x 3600 s

= 711.763 km

Basically, what you do to find the airplane's velocity with respect to the ground, is adding the vector of air velocity (with respect to ground), to the vector of the airplane's velocity (with respect to air). That's what zoeksyrianos did there. If you didn't understand the answer or need a more detailed answer, let me know.

I used Savio's method to find the speed of the plane with respect to the ground (which came out to be 175.14 m/s) but I couldn't understand how zoeksyrianos did it. I still can't figure out how to solve the rest

Perhaps this example might help


Answer:


Solutions also attached

Don't worry, I will look into it later today (probably at 15 hours from now) and answer all the questions.
Oh, and something else: "an angle of 30° west of due north" does that mean that the vector is closer to north, or closer to west? I ask because English isn't my native language and I am not sure if I understood the expression.

As I said earlier, I'm not sure what this expression means, so I had to assume. I hope I drawn the vector u2 correctly at the picture (attachment) otherwise everything is wrong!!!


Lets see: If u1 is the velocity of the airplane with respect to ground, u2 the velocity of the air with respect to ground, and u1,2 the velocity of the airplane with respect to air, we have: u1,2=u1-u2 <=> u1=u1,2+u2 (u1, u2 and u1,2 are vectors)

Adding vectors can be done with different ways, here is what I would do:
Firstly, break the u2 vector to u2x and u2y (see attachment). We have: u2x=sin30*u2*sin30=17,5 and u2y=u2*cos30=24,75
Add the u2x and the u1,2 (as they are both on the x axon): ux=u1,2+u2x=190-17,5=172,5m/s direction: east.
Now, we only have two vertical vectors that we need to add: ux and u2y. Because they are vertical, we use the formula:
u12=ux2+u2y2 <=>...<=>u1=174,27m/s, that's the speed
To find the heading: cos(ω)=u2y/ux=24,75/172,5=0,1435, so ω=81,75o

3) We will only take into account the component of u1 that is heading east. We already know that component: it's the ux=172,5. So, in 1 hour, the plane will travel east: x=172,5*3600=621.000m.
(again, I hope I haven't done anything wrong)

Thank you so much! That really helped