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Published bySofia Atkinson Modified over 3 years ago

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Graphical Analysis of Linear Motion

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A car travels along a road at a constant velocity of 10. m/s time (s) position (m) 0 0 1 10 2 20 3 30

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slope = Δx/Δt ΔtΔt ΔxΔx = 20 m/2 s = 10 m/s velocity

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position vs. time graph to find displacement: to find velocity: subtract values from graph find slope if straight line acceleration = 0

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Displacement for 1 st 3 seconds: 30 m area under graph slope:0acceleration

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v vs. t graph to find displacement: to find velocity: to find acceleration: find area read graph find slope

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Object dropped from a tall building (use g = 10 m/s 2 ) time (s) position (m) 0 0 1 5 2 20 3 45 x = x 0 + v 0 t + ½ at 2 x 0 = 0; v 0 = 0

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velocity @ 2 s: slope of tangent line slope = 20/1 = 20 m/s

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time (s) velocity (m/s) 0 0 1 10 2 20 3 30 v = v 0 + at

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acceleration @ 2 s:slope 10 m/s 2 displacement for 1 st 3 s:area Δx = 45 m

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area:30 m/s= Δv

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Graphs (vs. time) position velocityacceleration slope area

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If acceleration is positive (constant); x 0 = 0; v 0 = 0 x = ½ at 2

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To produce a straight-line graph: slope = ½ a also: t 2 vs. x, x vs. t, t vs. x

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Graphs of x, v, a

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