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Form 4 Phy & A. Math's
Relationship between v-t graph and s-t graph
Suppose a particle starts at the origin (i.e. s = 0 when t = 0).
If it travels at a constant velocity of 1.033 m/s (i.e. the velocity-time garph is a a line parallel to the x-axis), how can we determine the shape of the displacement-time graph?
When t = 0, s = 0, i.e. the particle has no dispalcement. Therefore a point (0, 0) is plotted on the second graph.
When t = 1.1, s = 1.033*1.1 = 1.136, which is the area of the rectangle formed at t = 1.1. Therefore a point (1.1, 1.136) is plotted on the second graph.
This process is repeated and the locus of points are formed on the second graph. It is a straight line with slope 1.033.
Suppose a particle starts at the origin (i.e. s = 0 when t = 0).
If it travels at with constantly increasing velocities (i.e. the velocity-time graph is a straight line with slope 0.474), how can we determine the shape of the displacement-time graph?
The average velocity at time t can be regarded as (velcoity at 0 + velocity at t)/2 = (0 + v)/2 = v/2 . The area of the triangle is the same as the area of the blue rectangle.
When t = 0, s = 0, i.e. the particle has no displacement. Therefore a point (0, 0) is plotted on the second graph.
When t = 1.436, s = (0.474*1.436/2)*1.436 = 0.489, which is the area of the triangle formed at t = 1.436. Therefore a point (1.436, 0.489) is plotted on the second graph.
This process is repeated and the locus of points are formed on the second graph. It is a curve which is part of a parabola.
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