Question

A person is standing 350 feet away from a model rocket that is fired straight up into the air at a rate of 15 feet per second. At what rate is the distance between the person and the rocket increasing 20 seconds after liftoff?

a) 15 feet per second
b) 10 feet per second
c) 5 feet per second
d) 0 feet per second

Answer 1

The rate at which the distance between the person and the rocket is increasing 20 seconds after liftoff is 0 feet per second.

Step-by-step explanation:

To find the rate at which the distance between the person and the rocket is increasing, we can use the concept of related rates. Let's call the distance between the person and the rocket 'x' and the time 't'. We are given that dx/dt (the rate at which x is changing) is 15 feet per second. We want to find d^2x/dt^2 (the rate at which dx/dt is changing) at t = 20 seconds.

We can start by finding dx/dt using the given information. dx/dt is the same as the rate at which x is changing, which is 15 feet per second.

Since dx/dt is constant, d^2x/dt^2 is zero. Therefore, the rate at which the distance between the person and the rocket is increasing 20 seconds after liftoff is 0 feet per second (option d).