Question

a port and a radar station are 2 mi apart on a straight shore running east and west. a ship leaves the port at noon traveling northeast at a rate of 15 mi/hr. if the ship maintains its speed and course, what is the rate of change of the tracking angle theta between the shore and the line between the radar station and the ship at 12:30 pm?

Answer 1

Answer:

The rate of change of the tracking angle is 0.05599 rad/sec

Step-by-step explanation:

Here the ship is traveling at 15 mi/hr north east and

Port to Radar station = 2 miles

Distance traveled by the ship in 30 minutes = 0.5 * 15 = 7.5 miles

Therefore the ship, port and radar makes a triangle with sides

2, 7.5 and x

The value of x is gotten from cosine rule as follows

x² = 2² + 7.5² - 2*2*7.5*cos(45) = 39.04

x = 6.25 miles

By sine rule we have

$\frac{sin A}{a} = \frac{sin B}{b}$

Therefore,

$\frac{sin 45}{6.25} = \frac{sin \alpha }{7.5}$

α = Angle between radar and ship α

∴ α = 58.052

Where we put

$\frac{sin 45}{6.25} = \frac{sin \alpha }{x}$ to get

$\frac{x}{6.25} = \frac{sin \alpha }{sin 45}$ and differentiate to get

$\frac{\frac{dx}{dt} }{6.25} = cos\alpha\frac{\frac{d\alpha }{dt} }{sin 45}$

$\frac{15sin45 }{6.25cos\alpha } =\frac{d\alpha }{dt} }$= 3.208 degrees/second = 0.05599 rad/sec.