5. Using the Laplace transform, we want to solve the second part of the initial value problem when the bungee jumper is

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5 Using The Laplace Transform We Want To Solve The Second Part Of The Initial Value Problem When The Bungee Jumper Is 1 (117.88 KiB) Viewed 36 times

5. Using the Laplace transform, we want to solve the second part of the initial value problem when the bungee jumper is 30 or more feet below the bridge. That is, we want to solve the following IVP using the Laplace Transform. mx2 + ax2-b(x2) = mg; for t > t₁ x2(t₁) = 0; x₂ (t₁) = v₁. Since the Laplace transform requires to know the value of x₂(t) at t = 0, we will define a new variable μ = t - t₁ and a new function y₂(μ) = x₂(μ+t₁). Notice that this is just applying a horizontal shift to x2, which will not change it's derivatives. Thus y2 would satisfy the same differential equation, but have the following initial conditions, my2 + ay + ky₂ = mg; y2 (0) = x₂(t₁) = 0; y₂ (0) = x₂(t₁) = V₁. We will solve this shifted initial value problem for y2(μ) using the Laplace transform, then apply y2(μ) = x₂(µ+t₁) = x2(t). Again, you may use a = 2.8 and g = 9.8, but leave m and k as unknown constants. The solution r2(t) represents your position below the natural length of the cord after it starts to pull back. (I recommend that you leave v₁, a, and g as variables when find the solution to the IVP, and only substitute the values of these three variables at the end.)