Question 4 A Solve the following initial-value problem. mx" + x' + kx = mg, X(t1) = 0, x'(t1) = v1 For now, you may use the value k = 14, but eventually you will need to replace that with the actual values for the cords you brought. The solution x(t) represents the position of your feet below the natural length of the cord after it starts to pull back. (You may use my = 160, = 1, and g = 32. Round your coefficients to three decimal places.) _ L x(t)= e (1) 52.19475 cos( Vgg)t+5.43746sin( V21(7)9)t+ 2 ft X Compute the derivative of the expression you found in question (4). (Round your coefficients to two decimal places.) x'(t) = ft/S Solve for the value of t (in s) where it is zero. This time is t2. Be careful that the time you compute is greater than tl there are several times when your motion stops at the top and bottom of your bounces! (Round your answer to two decimal places.) t2: 5 After you find t2, substitute it back into the solution you found in question (4) to nd your lowest position (in ft below the bridge). (Round your answer to one decimal place.) x02) = x ft