Question 7fb29

May 20, 2016

Skydiver is accelerating, increasing the air resistance due to bigger speed, therefore reducing the acceleration as he descends, until the point of terminal velocity, where the speed is the maximum and the acceleration is 0 due to the air resistance being equal to the gravitational force.

Explanation:

As the skydiver descends, two forces are acted upon him. Gravity ${F}_{g}$ and air resistance ${F}_{r e s}$. What connects these with the acceleration is Newton's 2nd law:

ΣF=m*a

Where Σ# notes the sum of all forces. In this case, noting the downwards force as positive:

${F}_{g} - {F}_{r e s} = m \cdot a$

Since you are interested in $a$, solving with respect to it:

$a = \frac{{F}_{g} - {F}_{r e s}}{m}$ $\left(E q u a t i o n 1\right)$

We can assume that the height is small enough so that the force of gravity does not change. Also, the mass of the skydiver does not change. This means that acceleration depends only on the resistance of the air, which is not constant. It actually depends on the speed of the skydiver, since the faster the skydiver is, the stronger he pushes the air, so the air pushes him back (resists) stronger, which is expressed by Newton's 1st law.

Therefore, we know that as the velocity of the skydiver increases, the resistance increases as well and due to equation 1 as resistant force increases acceleration decreases. The term terminal velocity refers to the point at which the forces of gravity and resistance are equal, so due to equation 1 acceleration is set to zero and the object can not increase its speed anymore.