A hot-air balloon is 150 feet above the ground when a motorcycle passes directly beneath it (traveling in a straight line on a horizontal road). The motorcycle is traveling at 40 mph. If the balloon is rising vertically at a rate of 10 ft/sec, what is the rate of change of the distance between the balloon and the motorcycle ten seconds later?

Question

Physics

Sydney

Today, 07:46

A hot-air balloon is 150 feet above the ground when a motorcycle passes directly beneath it (traveling in a straight line on a horizontal road). The motorcycle is traveling at 40 mph. If the balloon is rising vertically at a rate of 10 ft/sec, what is the rate of change of the distance between the balloon and the motorcycle ten seconds later?

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Dylan Herrera · Answer 1

As you know that

a^2+b^2=c^2

2da/dt+2bdb/dt=2cdc/dt

change the units into ft/sec or miles per hour

so your da is the balloon rate, your db is the motorcycle rate. a is 150 feet,

b is 40 m/h x 10 secs, just plug it in to figure out c first. then dc/dt.