Final answer:
Initially, the skydiver experiences accelerated motion due to gravity (freefall) until the parachute is deployed, after which she reaches terminal velocity and experiences zero acceleration. A kinematic equation V = at is used to calculate velocity before opening the parachute, with acceleration being 9.80 m/s². The acceleration-time graph shows a constant 9.80 m/s² during freefall that decreases to near zero upon parachute deployment. option B is correct
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Step-by-step explanation:
When a skydiver jumps out of a plane, their motion can option.be described in several phases:
For the first 10 seconds, the skydiver is in freefall motion, accelerating due to the force of gravity. This is an example of accelerated motion where the acceleration is the acceleration due to gravity, commonly denoted as g and having a value of approximately 9.80 m/s² downward.
Upon opening her parachute, the upward force of air resistance matches the downward force of gravity creating a state of zero acceleration motion. This means the skydiver has reached what is termed 'terminal velocity', where her speed remains constant.
At landing, there's a brief period of deceleration until the skydiver comes to a stop.
Regarding the kinematic equation to use for finding velocity after 2 seconds before deploying the parachute, the appropriate equation is V = at, with V being the final velocity, a the acceleration due to gravity (9.80 m/s²), and t the time in seconds. Hence, V = 9.80 m/s² × 2s = 19.6 m/s.
The shape of the graph of the magnitude of acceleration versus time for a falling skydiver would show a constant acceleration of 9.80 m/s² during freefall, decreasing to nearly zero once the parachute is deployed and the skydiver has reached terminal velocity.