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Step by Step solution anyone?
14
Ok so i've got the question, the working and the answer but i don't really understand it so was hoping someone could help me out.
The question is.
An airplane is flying at a constant speed at a constantly altitude of 3km in a straight line that will take it directly over an observer at ground level. At a given instant, the observer notes that the angle theta is 1/3 (pi) radians and is increasing at 1/60 radians per second. Find the speed in km/h, at which the airplane is moving towards the observer.
The question is.
An airplane is flying at a constant speed at a constantly altitude of 3km in a straight line that will take it directly over an observer at ground level. At a given instant, the observer notes that the angle theta is 1/3 (pi) radians and is increasing at 1/60 radians per second. Find the speed in km/h, at which the airplane is moving towards the observer.
Reply 1
try drawing a diagram
1) Ok the way id do it is to use trig to work out the disfference in distance in one second.
2) you can then just convert this in kmh^-1.
however one thing bugs me - the plane is flying 3000m above ground level but the person isnt viewing the plane from ground level unless they are lieing down... i know this is silly as ur just making an approximation but i think i might be wrong becaseu of this. So im not going to tell you my answer and my working as i think its wrong or at least there is a better way of doing it.
1) Ok the way id do it is to use trig to work out the disfference in distance in one second.
2) you can then just convert this in kmh^-1.
however one thing bugs me - the plane is flying 3000m above ground level but the person isnt viewing the plane from ground level unless they are lieing down... i know this is silly as ur just making an approximation but i think i might be wrong becaseu of this. So im not going to tell you my answer and my working as i think its wrong or at least there is a better way of doing it.
Reply 2
Ok, the best way to think about it is to draw an imaginary right-angled triangle between the observer, the plane and the ground. I'll assume you mean theta is the angle made as the observer looks up at the plane from the ground. This means that the angle with the vertical is π/2-π/3 = π/6
..Plane ------X
.......\.......|_|
........\.........|
.........\........|
..........\.......|3km
...........\.π/6.|
............\.....|
.............\....|
______Observer
Now we can work out the distance that the plane is from the point directly above the observer, that I've called X. This is 3tan(π/6) = √3 km
Now here's the clever bit. You could try working out the distance the plane travels in a second but then you'll have to work out the tan of some horrible angle. What you can instead do is notice that if the plane covers 1/60 rad per second, then in 10π seconds it would have covered 10π/60 = π/6 rad.
Thus, the speed of the plane is √3 ÷ 10π = 0.055... km/sec
= 198 km/h (to 3sf)
Check that one time
..Plane ------X
.......\.......|_|
........\.........|
.........\........|
..........\.......|3km
...........\.π/6.|
............\.....|
.............\....|
______Observer
Now we can work out the distance that the plane is from the point directly above the observer, that I've called X. This is 3tan(π/6) = √3 km
Now here's the clever bit. You could try working out the distance the plane travels in a second but then you'll have to work out the tan of some horrible angle. What you can instead do is notice that if the plane covers 1/60 rad per second, then in 10π seconds it would have covered 10π/60 = π/6 rad.
Thus, the speed of the plane is √3 ÷ 10π = 0.055... km/sec
= 198 km/h (to 3sf)
Check that one time
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