Motion detector graphs, distance vs time

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SUMMARY

The discussion centers on the graphical representation of distance versus time for objects in motion, specifically a ball dropped from a height of 1 meter and a car parked on a steep hill. It is established that as time progresses, the downward speed of both the ball and the car increases, resulting in a non-linear distance-time graph. The correct representation for the ball's motion is a curved graph that rises to the right, indicating increasing distance over time. A straight line would not accurately depict the acceleration of the objects involved.

PREREQUISITES
  • Understanding of basic physics concepts such as acceleration and free fall
  • Familiarity with graphing distance versus time
  • Knowledge of the kinematic equations of motion
  • Ability to interpret motion graphs
NEXT STEPS
  • Study the kinematic equations for uniformly accelerated motion
  • Learn about the graphical representation of motion in physics
  • Explore the concept of acceleration due to gravity and its effects on falling objects
  • Investigate the differences between linear and non-linear graphs in motion analysis
USEFUL FOR

Students studying physics, educators teaching motion concepts, and anyone interested in understanding the graphical analysis of motion in real-world scenarios.

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hi!
a ball is dropped from a height of 1 meter above the floor. Take the origin to be the point from which the ball is released.
would the graph be like a curved graph rising to the right starting from the origin?

how would the graph look like if a car is parked on a steep hill?

thank you
 
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As time progresses the downwards speed of the ball (and the car) increases. If the speed of an object increases it covers more distance in the same time. This implies that the object covers more distance in the same time interval at a later time in its motion. How should the distance - time graph have to look then? Will a straight upwards sloping graph do?
 
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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