A skydiver is 760 meters above the ground when he opens his parachute.

After opening the parachute, he descends at a constant speed of 17

meters per second. Part A: What type of function would best model this

situation?

s (t) = 760 - 17*t

Step-by-step explanation:

Solution:-

- We will denote a varable (s) : The distance of the sky-diver from the ground.

- The initial distance of the sky-diver from the ground when he opened his parachute was (s_o = 760 meters).

- After opening his parachute he experiences air resistance due to enlarged surface area of the parachute. This sudden rush of air entrapped under the parachute induces an upward force called drag force.

- The drag force (D) opposes the downward force due to gravity i. e weight (W) until the two forces balances and the net acceleration (a) of the sky-diver is zero.

- This is expressed by an equation of motion:

W - D = m*a

Where,

m: Mass of sky-diver

When,

W = D, a = 0

- The term constant speed or terminal velocity (v) is achieved when the above condition (W = D & a = 0) is achieved.

- We are given that the sky-diver desceneds down at a constant speed of 17 m/s. The direction of motion is opposite to the direction of displacement (s) of the sky-diver from the ground. Hence, the terminal velocity of the sky-diver is (v = - 17 m/s).

- We can use kinematic equation of motions to express the displacement of sky-diver from the ground as a function of time (t):

s (t) = s_o + v*t

s (t) = 760 - 17*t

- The above is a linear function of displacement (s) of the sky-diver from the ground of time (t) after he opens his parachute.