Answer:
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
