3. A man stands 200 ft to the left of where a car starts moving north. The car is
moving at 50ft/sec. (a) Find the rate at which the diagonal distance from the man
to the car is changing once the car has been traveling for 10 seconds. (b) Find the
rate at which the angle from the man to the car is changing at the same moment
(10 seconds after the car starts moving). Round answers to 4 decimal places.
200 ft
car (moving at 50 ft/sec)
O

3 A man stands 200 ft to the left of where a car starts moving north The car is moving at 50ftsec a Find the rate at which the diagonal distance from the man to class=

Respuesta :

The length of the diagonal will be 538.5 ft. Then the rate at which the angle from the man to the car is changing at the same moment will be 46.43 ft/s.

What is speed?

The distance covered by the particle or the body in an hour is called speed. It is a scalar quantity. It is the ratio of distance to time.

We know that the speed formula

Speed = Distance/Time

A man stands 200 ft to the left of where a car starts moving north. The car is moving at 50ft/sec.

The rate at which the diagonal distance from the man to the car is changing once the car has been traveling for 10 seconds. Then the length of the diagonal will be

L² = x² + y²

L² = 200² + (50 × 10)²

L² = 40000 + 250000

L² = 290000

L = 538.50 ft

Then the rate at which the angle from the man to the car is changing at the same moment will be

L² = x² + y²

Differentiate with respect to time, then we have

         2L (dL/dt) = 0 + 2y (dy/dt)

2(538.5) (dL/dt) = 2 (500) (50)

                dL/dt = 46.43 ft/sec

More about the speed link is given below.

https://brainly.com/question/7359669

#SPJ1

ACCESS MORE