Speed of a plane given distance and time equation.

[brandy]

an aircraft at an altitude of 10 000 m is flying at a constant speed on a line that will take it directly over an observer on the ground. if the observer notes that the angle of elevation o the aircraft is 60 degrees and is increasing at a rate of 1 degree per second find the speed of the plane in metres per second.

i got the formula for the equation as 10 000/(Tan (a+x))=m
a=angle x=seconds m=metres across.

i can't think why it would be incorrect. the velocity is metres per second so delta M over Delta T or Delta y over Delta x (the gradient function)
that didnt get me anywhere because i didnt know wat to sub in for the equation

then i simply took an m value and a x value and used the average velocity formula but as expected it changed.

plllllllllleeeassse i know the answer is staring me right in the face, I am just temporaily blind. help me.

 

[tiny-tim]

Hi brandy!

(have a delta: ∆ )

i got the formula for the equation as 10 000/(Tan (a+x))=m
a=angle x=seconds m=metres across.

hmm … your notation is horrible, and you've left out half of the rhs …

you should have ∆x = v ∆t = 10 000/(Tan (60º + ∆t)) - 10 000/(Tan (60º))

 

[Architeuthis Dux]

I would recommend using the distance formula, which is d = rt, where d is the distance, r is the rate or speed, and t is the time. In this case, we know the distance (10,000 meters) and the time (1 second per degree), so we just need to solve for the speed (r).

First, we need to convert the angle from degrees to radians, as that is the unit used in trigonometric functions. We can do this by multiplying 60 degrees by π/180. This gives us an angle of π/3 radians.

Next, we can use the tangent function to find the rate. Tan(π/3) = opposite/adjacent = 10,000/r. Solving for r, we get r = 10,000/tan(π/3) = 10,000/√3 ≈ 5,774 meters per second.

So, the speed of the plane is approximately 5,774 meters per second. Hope this helps!