Help with a VT Graph: Acceleration & Deceleration Explained

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The discussion revolves around creating a velocity-time (VT) graph for a parachute jump, focusing on the phases of acceleration and deceleration. During the initial phase (0-2 seconds), the jumper accelerates uniformly. From 2-4 seconds, the jumper reaches a constant velocity, and from 4-6 seconds, the graph shows a decreasing gradient as the jumper decelerates until reaching zero velocity. The conversation also raises questions about the effects of parachute deployment on descent speed and impact. Understanding these phases is crucial for accurately representing the jump in the VT graph.
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I need to do a VT graph showing a parachute jump, I need an explanation about the graph too.

Homework Equations


None

My Attempt:
Fhsst_rectmot15.png


From point 0-2 the jump is accelerating largely at a constant uniform. From point 2-4 the acceleration remains at a constant velocity. From point 4-6 the gradient decreases at a constant speed until it reaches zero.

Thanks a lot,
GCSEsoon
 
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GCSEsoon said:
I need to do a VT graph showing a parachute jump, I need an explanation about the graph too.

Homework Equations


None

My Attempt:Thanks a lot,
GCSEsoon

What is it you are attempting to show with your graph. Perhaps a little more of an explanation of your thinking about your graph is in order?
 
From point 0-2 the jump is accelerating largely at a constant uniform. From point 2-4 the acceleration remains at a constant velocity. From point 4-6 the gradient decreases at a constant speed until it reaches zero.
 
GCSEsoon said:
From point 0-2 the jump is accelerating largely at a constant uniform. From point 2-4 the acceleration remains at a constant velocity. From point 4-6 the gradient decreases at a constant speed until it reaches zero.

What happens when the chute deploys? The diver stops at constant velocity or slows to a final velocity of descent?

At the ground, how fast will the jumper stop? 2 seconds to hit the ground and slow his impact?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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