guys can anyone please help me with this word problem from my

tan(45°) = 8/p  Hence p = 8/tan(45°)  or p = 8 m

cos(45°) = (w+p)/27  or  1/√2 = (w + 8)/27 so  w = 27/√2 - 8 m

$${\mathtt{w}} = {\frac{{\mathtt{27}}}{{\sqrt{{\mathtt{2}}}}}}{\mathtt{\,-\,}}{\mathtt{8}} \Rightarrow {\mathtt{w}} = {\mathtt{11.091\: \!883\: \!092\: \!036\: \!783\: \!2}}$$

or w ≈ 11.1 m

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problem1:

Let angle of elevation at any time be θ and the horizontal distance to the plane be h.  Then:

tanθ = 440/h

Differentiate both sides with respect to time, t.

(tan²θ + 1)dθ/dt = -440/h²*dh/dt

The plane's speed is just dh/dt (let's call it v), so v = -(tan²θ + 1)h²*dθ/dt/440 

We are told that when θ = 45°  dθ/dt = -0.05radians/sec.  and, since tan(45°) = 1, then h = 440m, so it's just a case of plugging in the numbers:

$${\mathtt{v}} = {\frac{\left({{\mathtt{1}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}{{\mathtt{440}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{0.05}}}{{\mathtt{440}}}} \Rightarrow {\mathtt{v}} = {\mathtt{44}}$$

v = 44 m/s

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Why not show us what you did for problem 2 and then we can comment on it?

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For question 2, consider this:

The girder must rotate a full 90 degrees (Or pi/4 if you like your radians), so at one point or another it's going to be at 45° to the corridor walls. Seeing as we ignore width of the beam we can forget about it 'catching' on the walls when it rotates.

So we have a nice situation of all we need is the corridor corner to support a three-quarter circle of diameter 27 metres. This is a radius of a 27/2 = 13.5 metres. So the corridor needs to be 13.5 metres wide.

If my explanation is a bit confusing, just say and I will try to mock up some diagrams to better explain it.

Thank you Alan and Sir-Emo, 

I intend to examine your answers in detail a little later.  

I am very curious about this one :)

I completely skimmed over the detail the first corridor is only 8m wide -_- I thought they were the same width. Alan has the right idea.

Here's another way to do this based on Alan's drawing...

The hypotenuse of the the small triangle at the bottom left = √[8^2 + 8^2] = 8√2

So, the length of the hypotenuse of larger triangle at the top right is just 27 - 8√2

And using similar triangles, the length of each side of this larger triangle = w =  [27 - 8√2]/ √2  = about 11.092 ft.  ≈  11.1 ft

@Alan @Sir Emo @CPhill,guys sorry for the late reply, and yeah i also arrive at the same aswer, mine is 11.098 to be exact, my solution is a little bit longer(though it's pretty much the same as your's) and i dont have a pc to write it down like you guys (im using my mom's ipad) so it might confuse you guys if i wright it here manually

Hi Alan,

I have finally gotten back to this one.

I have been looking at the aeroplane one.  I can certainly see how you got an airspeed of 44m/s   

I am wondering though.  

If you had been asked what is the speed at which the plane is moving away from the observer at that point, what would be the answer?

I don't understand what you are asking Melody; that is the speed at which the plane is moving away from the observer!

Or perhaps you are asking how it would be done if the plane were immediately above the observer.  In this case, take a small interval of time δt during which the angle changes by δθ.  Now we have  tan(δθ) = vδt/440.

When δθ is small, tan(δθ) = δθ, so δθ = vδt/440  or v = 440*δθ/δt.

Given the rate of change of angle, δθ/δt, we get the plane's speed.

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Thanks Alan,

Mmm, I did wonder if my question made sense :/

The plane is flying at a horizontal speed of 44m/s.  And a vertical speed of 0m/s

But the observer is not at the same altitude as the plane so I thought the speed from the observers view point would be different  ://

Umm.   The observer sees this 'speed' in radians per second.  I want the change in 'site' distance (hypotenuse) per second.

Does that make sense or am I makeing no sense at all ?

Ok.  In that case just replace tanθ = 440/h  by sinθ = 440/h where h is now the line of sight distance.

cosθ*dθ/dt = -440/h²dh/dt    when θ = 45°  then cosθ = 1/√2,  h = 440*√2 so

dh/dt = 440^2*2*(1/√2)*0.05/440 = (22*√2) m/s ≈ 31.1 m/s

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Of course, thanks Alan, that is what I was asking :)