Question

an airplane headed 30 degrees northeast climbing on a 20 degrees incline at a speed of 240 miles per hour passes 1 mile directly over a car headed south on a flat road at the rate of 60 miles per hour at what rate of speed are they separating after 10 minutes?

Answer 1

An airplane headed 30 degrees northeast climbing on a 20 degrees incline at a speed of 240 miles per hour passes 1 mile directly over a car headed south on a flat road at the rate of 60 miles per hour at what rate of speed are they separating after 10 minutes?

Solution:

$B_{1}$ - the point at which a car was when an airplane was flying over it

$A_{1}$ - the point at which there was an airplane as it flew over the car

$B_{2}$ - the point at which a car was after $t$ hours

$A_{2}$ - the point at which an airplane was after $t$ hours

$S = A_{2}B_{2}$ - the distance between an airplane and a car after $t$ hours

$A_{1}B_{1} = 1ml$

$B_{1}B_{2} = 60t$

$A_{1}A_{2} = 240t$

$HA_{2} = 1 + A_{1}A_{2}\sin 20{}^{\circ} = 1 + 240\sin 20{}^{\circ}t = 1 + 82t$

$HB_{2}^{2} = B_{1}B_{2}^{2} + B_{1}H^{2} - 2\cdot B_{1}B_{2}\cdot B_{1}H\cos 150{}^{\circ}$

$B_{1}H = A_{1}A_{3} = A_{1}A_{2}\cos 20{}^{\circ} = 240\cos 20{}^{\circ}t = 226t$

$HB_{2}^{2} = (60t)^{2} + (226t)^{2} - 2\cdot 60t\cdot 226t\cos 150{}^{\circ} = 78,163t^{2}$

$S^{2} = A_{2}B_{2}^{2} = HB_{2}^{2} + HA_{2}^{2} = 78,163t^{2} + (1 + 82t)^{2} = 84,887t^{2} + 164t + 1$

$S(t) = \sqrt{84,887t^2 + 164t + 1}$

The rate of speed is:

$$v (t) = \frac {d S}{d t} = \frac {1 6 9 , 7 7 4 t + 1 6 4}{2 \sqrt {8 4 , 8 8 7 t ^ {2} + 1 6 4 t + 1}}$$

When $t = 10 \, \text{min} = \frac{1}{6} \, \text{hour}$

$$v \left( \frac{1}{6} \right) = \frac{ \frac{169,774}{6} + 164 }{ 2 \sqrt{ \frac{84,887}{36} } + \frac{164}{6} + 1 } = 291 \, \text{ml/hour}$$