How can we derive the distance formula using velocity and acceleration?

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SUMMARY

The distance formula S = vt + 1/2at² can be derived from the differential equation d²s/dt² = C, where C is a constant representing acceleration. By analyzing a velocity-time graph for an object under constant acceleration, the area of the trapezium formed provides the displacement. This graphical representation confirms the formula's validity, illustrating how velocity and acceleration contribute to calculating distance.

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Homework Statement



I try to drive distance formula which is S=vt+1/2at^2.



I use velocity and acceleration formula to drive the distance formula but i couldn't get the distance formula.is there anyone know how to drive S=vt+1/2at^2. and how that came from in the first place .

thank you.
 
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Its the solution to the differential equation:

\frac{d^2s}{dt^2} = C

where c is a constant.
 
A more graphical method involves drawing a velocity-time graph of an object under constant acceleration from velocity u to v. It should look like a trapezium whose parallel sides are parallel to the velocity axis. The area of the parallelogram gives you the displacement, and that is given by the formula.
 
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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