Final answer:
Skydivers initially experience acceleration at 9.8 m/s² downwards due to gravity. As they continue to fall, air resistance increases until it equals the force of gravity, thereby reducing the acceleration gradually to zero when they reach terminal velocity.
Step-by-step explanation:
When considering the acceleration of skydivers, it's important to recognize that they initially accelerate under the force of gravity at 9.8 m/s² downwards. Once they jump from a plane, for the first couple of seconds, they would indeed experience this acceleration. Using the kinematic equation V = at, if we assume a skydiver has been falling for 2 seconds, the velocity (V) would be calculated as the acceleration due to gravity (a) times the time (t), giving us V = 9.8 m/s² × 2 s = 19.6 m/s.
However, as they fall, skydivers encounter air resistance, which increases with their speed until it equals the force of gravity. This leads to a state called terminal velocity, where the net acceleration is zero (0 m/s²) because the upward force of the air resistance is equal to the downward force of gravity. The graph of acceleration versus time for a falling skydiver would show a steep downward slope at first, indicating the increase in acceleration due to gravity. As the skydiver falls longer, the slope gradually approaches zero as they reaches terminal velocity.
Considering a downward direction as positive, we would say that a skydiver initially has an acceleration of +9.8 m/s². If considering upward as positive, this acceleration would be -9.8 m/s². Once the skydiver reaches terminal velocity, the acceleration would be 0 m/s² irrespective of the coordinate system orientation.