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HallsofIvy

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8(0.05-x)- 0.32= 0.08- 8x for x< 0.05

= -0.032 for 0.05< x< 0.15

Since "force= mass* acceleration", we have the differential equation

0.0053x"= 0.08- 8x for x< 0.05

= -0.032 for 0.05< x< 0.15

with initial conditions x(0)= 0, x'(0)= 0.

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