A plane flying with a constant speed of 390 km/h passes over a ground radar station at an altitude of 3 km and climbs at an angle of 30°. At what rate is the distance from the plane to the radar station increasing a minute later? (Round your answer to the nearest whole number.) 325 Incorrect: Your answer is incorrect. km/h

The answer to the question is;

One minute later the distance from the plane to the radar station is increasing at a rate of 387.049 kph

Explanation:

Spped of plane = 390 km/h = 108.3 m/s

angle of 30 degrees vertical component of velocity = 390 * sin 30 = 195 km/h = 6.45 km/min

after one minute we have

6.45 km

Using cosine rule we have

a^2=b^2+c^2-2bc cosA

Where A = 120 and

b = Vertical height of plane above radar = 1 km

a = Distance of plane from radar

c = Distance moved by plane in one minute = third side of triangle abc

Solving with b = 1 kph gives

a^2=1^2+c^2-2bc cos(120)

= 1 + c^2-2×1×c ×(-1/2)

=1+c^2+c or

a² = 1 + c² + c

Where c = 6.5 a =7.0533

To find the rate of change of a with time, we have

2a(da/dt)=0+2c(dc/dt)+dc/dt

(da/dt) = (2c+1)(dc/dt)/(2a)

Which gives

(da/dt)= (2×6.5+1)×6.5÷(2×7.0533)

= 6.45km/min

Multiply by 60 min/hour, we have

6.45km/min×60 min/hour

=387.049 kph

pe = weightxheight = 2000x1000=2,000,000joules

need to see circuit to get time constant