Question

2. A spy uses a telescope to track a rocket launched vertically from a launch pad 6 km away. The rocket travels upwards at a velocity of 21.6 km per minute. Find the rate of change of the distance between the spy and the rocket when the rocket is 8 km in the air. Answer: 17.28 km/mim

NEED HELP ASAP

Answer 1

The rate of change of the distance between the spy and the rocket when the rocket is 8 km in the air is 17.28 km/min.

Step-by-step explanation:

To find the rate of change of the distance between the spy and the rocket, we can use the concept of related rates.

Let's consider the distance between the spy and the rocket at any given time as d, and the height of the rocket at that time as h.

We know that the rocket travels upwards with a velocity of 21.6 km per minute. This means that the rate at which the height of the rocket is changing is 21.6 km/min.

Using the Pythagorean theorem for right triangles, we can relate the distance and height:

d^2 = 6^2 + h^2

Differentiating both sides with respect to time, we get:

2d(dd/dt) = 2h(dh/dt)

Let's substitute the values we know: when the rocket is 8 km in the air, we have d = 6 km and h = 8 km. We also know that dh/dt (rate at which the height is changing) is 21.6 km/min.

Now, we can solve for d(dt/dd) to find the rate of change of the distance:

d(dt/dd) = (2h(dh/dt))/(2d)

Substituting the values, we have: d(dt/dd) = (2 * 8 * 21.6)/(2 * 6) = 17.28 km/min

Answer 2

The rate of change in distance between the spy and the rocket is 10.8 km/min,

or about 180 m/sec.

Step-by-step explanation:

At "Time=0" the rocket is 6km from the spy.

The rocket travels vertically 8km in about 0.37 minutes, calculated by 8km÷21.6km/min.

At that point, the rocket is 10 km from the spy., calculated by 6²+8²=10²

The difference is 4 kiilometers over 0.37minutes, about 22.2 seconds

Divide Distance by Time to get Rate:

4km/0.37min = 10.8 km/min

That is about 180 meters per second.