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A road running north to south crosses a road going east to west at the point P. Car A is driving north along the first road, and an airplane is flying east above the second road. At a particular time the car is 15 kilometers to the north of P and traveling at 60 km/hr, while the airplane is flying at speed 185 km/hr 10 kilometers east of P at an altitude of 2 km. How fast is the distance between the car and the airplane changing?

Answer :

CandyBbygo

Answer:

D^2 = (x^2 + y^2) + z^2

 

and taking derivative of each term with respect to t or time, therefore:

 

2*D*dD/dt  =  2*x*dx/dt  + 2*y*dy/dt  + 0 (since z is constant)

 

divide by 2 on both sides,

 

D*dD/dt = x*dx/dt + y*dy/dt

 

Need to solve for D at t =0,  x (at t = 0) = 10 km,  y (at t = 0) = 15 km

at t =0,

 

D^2 =  c^2 + z^2 = (x^2 + y^2) + z^2  =  10^2 + 15^2 + 2^2  = 100 + 225 + 4  = 329

 

D = sqrt(329)

 

Therefore solving for dD/dt, which is the distance rate between the car and plane at t = 0

 

dD/dt = (x*dx/dt + y*dy/dt)/D  =  (10*190 + 15*60)/sqrt(329)  = (1900 + 900)/sqrt(329)

 

= 2800/sqrt(329) = 154.4 km/hr

 

154.4 km/hr

Step-by-step explanation:

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