A linear function because there is a constant rate change.
The required equation would best model is [tex]y = 760-17x[/tex].
Given that,
A skydiver is 760 meters above the ground.
After opening the parachute, he descends at a constant speed of 17 meters per second.
We have to determine,
What type of function would best model this situation and the equation.
According to the question,
Let, The distance of the sky-diver from the ground is s.
And the initial distance of the sky-diver from the ground when he opened his parachute [tex]= s_0 = 760m[/tex]
- 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 opposes the downward force due to gravity .
Weight ( W ) until the two forces balances and the net acceleration (a) of the sky-diver is zero.
The equation is expressed by an equation of motion:
W = ma
Where, m = mass of skydiver.
The term constant speed or terminal velocity is achieved when the above condition ( W = D & a = 0 ) is achieved.
The sky-diver descends down at a constant speed of 17 m/s.
The direction of motion is opposite to the direction of displacement (y) of the sky-diver from the ground.
Hence, The terminal velocity of the sky-diver is (v = - 17 m/s)
A linear function because there is a constant rate change.
- By using kinematic equation of motions to express the displacement of sky-diver from the ground as a function of time(x):
[tex]y = 760-17x[/tex]
The above is a linear function of displacement (y) of the sky-diver from the ground of time (x) after he opens his parachute.
Hence, The required equation would best model is [tex]y = 760-17x[/tex].
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