proof -force is directly proportional to the square of the distance
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You should be familiar with the formula s=ut+12at2s=ut+12at2 which is one of the SUVAT equations for uniform motion which relate distance ss, initial and final velocities uu and vv, acceleration aaand time tt.
If s=0s=0 at t=0t=0 then u=0u=0. If aa is constant then s=12at2s=12at2 so plotting ss vs t2t2 should give you a straight line of the form y=mxy=mx going through the origin. The slope is m=12am=12a.
Possibly there is uncertainty in your data points, so that they are spread randomly around the line y=mxy=mx.
More likely is that s≠0s≠0 at t=0t=0. Then the relationship between ssand t2t2 is more complicated. The easiest option is probably to write the equation as
s=12a(t2+2uat+u2a2)−u22a=12a(t+ua)2−u22as=12a(t2+2uat+u2a2)−u22a=12a(t+ua)2−u22a
which is of the straight-line form y=mx+cy=mx+c if x=(t+ua)2x=(t+ua)2.
To obtain a straight line you need to add or subtract an amount t0t0 to your values of tt and adjust this value until the R2R2 value of your trend line is as close as possible to 1. This is quite easy to do in a spreadsheet. The amount t0t0which maximises R2R2 gives you a value for t0=uat0=ua. As a check, the y-intercept on your graph should be c=−u22a=−mt20c=−u22a=−mt02 where m=12am=12a is the slope of your graph.
If s=0s=0 at t=0t=0 then u=0u=0. If aa is constant then s=12at2s=12at2 so plotting ss vs t2t2 should give you a straight line of the form y=mxy=mx going through the origin. The slope is m=12am=12a.
Possibly there is uncertainty in your data points, so that they are spread randomly around the line y=mxy=mx.
More likely is that s≠0s≠0 at t=0t=0. Then the relationship between ssand t2t2 is more complicated. The easiest option is probably to write the equation as
s=12a(t2+2uat+u2a2)−u22a=12a(t+ua)2−u22as=12a(t2+2uat+u2a2)−u22a=12a(t+ua)2−u22a
which is of the straight-line form y=mx+cy=mx+c if x=(t+ua)2x=(t+ua)2.
To obtain a straight line you need to add or subtract an amount t0t0 to your values of tt and adjust this value until the R2R2 value of your trend line is as close as possible to 1. This is quite easy to do in a spreadsheet. The amount t0t0which maximises R2R2 gives you a value for t0=uat0=ua. As a check, the y-intercept on your graph should be c=−u22a=−mt20c=−u22a=−mt02 where m=12am=12a is the slope of your graph.
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