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Graphical Analysis of Linear Motion
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
slope = Δx/Δt ΔtΔt ΔxΔx = 20 m/2 s = 10 m/s velocity
position vs. time graph to find displacement: to find velocity: subtract values from graph find slope if straight line acceleration = 0
Displacement for 1 st 3 seconds: 30 m area under graph slope:0acceleration
v vs. t graph to find displacement: to find velocity: to find acceleration: find area read graph find slope
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
velocity @ 2 s: slope of tangent line slope = 20/1 = 20 m/s
time (s) velocity (m/s) 0 0 1 10 2 20 3 30 v = v 0 + at
acceleration @ 2 s:slope 10 m/s 2 displacement for 1 st 3 s:area Δx = 45 m
area:30 m/s= Δv
Graphs (vs. time) position velocityacceleration slope area
If acceleration is positive (constant); x 0 = 0; v 0 = 0 x = ½ at 2
To produce a straight-line graph: slope = ½ a also: t 2 vs. x, x vs. t, t vs. x
Graphs of x, v, a
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