A hot air balloon is 170 ft above the ground when a motorcycle (traveling in a s

Question

A hot air balloon is 170 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 45 mi / hr (66 ft / s). If the balloon rises vertically at a rate of 12 ft/s, what is the rate of change of the distance between the motor cycle and the balloon 10 seconds later?

A port and a radar station are 3 mi apart on a straight shore running east and west. A ship leaves the port at noon traveling 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

between the shore and the line between the radar station and the ship at 12:30 PM? (Hint: use law of sines)

Use the Law of Sines to find an equation relating the angle, ,the angle that the ship left the port at, the distance between the radar system and the ship, a, and the distance between the port and the ship, s. Evaluate any known trigonometric functions of 45degrees° as needed. Use radians to express any other angles.

Explanation / Answer

1.Let t sec be the time since the balloon was directly above the bike, and let d ft be the distance between the balloon and the bike.

According to problem,
d^2 = (66t)^2 + (170 + 12t)^2
= 4356t^2 + 28900 + 4080t + 144t^2
= 4500t^2 + 4080t+28900
2d dd/dt = 9000t + 4080
dd/dt = (4500t + 2040)/d

Now according to the problem ,

d^2= (660)^2+(170+120)^2 [ since t=10]

or , d=720.9022125

so, dd/dt = (4500t + 2040)/d =(45000+2040)/720.9022125=65.25 ft/sec

the rate of change of the distance between the motor cycle and the balloon 10 seconds later is 65.25 ft/sec.

2.

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