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By seeing the basic deifinition v=ds/st and a=dv/dt, so for a free fall of a body the acceleration of the body remains constant its graph will straight line parallel to the time axis, velocity graph will be one degree higher as it is obtained by integrating acceleration, hence its graph will be one degree higher and will be inclined at some angle and the graph of displacement will be still one degree higher (i.e. parabola)

notes: 1) area under the velocity time diagram for a given time interval is equal to the displacement over the interval 2) similar area under acc time diagram gives velocity over that period.


On earth, the acceleration due to gravity is a constant 9.8 m/s2 so a graph of acceleration vs time would have the equation a = 9.8 m/s2. As said above, the graph of acc. vs t would be a horizontal line (zero slope) with a y-intercept of 9.8 m/s2. The area under this line is a rectangle with a width, t, and a height of 9.8 m/s2, so change in velocity is given by delta v = (9.8m/s2) t. The area under this line is a triangle with a height of delta v and a base of t, so the equation for change in position or displacement is delta x = 1/2 (delta v)(t) or delta x = (4.9 m/s2)t2


If you know the equation of the displacement vs time graph for freefall (delta x = 1/2 g t2 where g=9,8 m/s2), get the graph by plotting delta x for various times. From that graph, find the slope of the tangent to the curve (A TI-83 calculator can do this) at various times. Plotting slope of tangent vs time gives you a velocity vs time graph. Plotting the slope of the velocity/time graph vs time gives you the acceleration vs time graph.

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Q: How do you construct position graph velocity and graph and acceleration timetables and graph all three cases of free fall?
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