Using the formula giving calculate the data

\f10:47PM Wed May 15 X Mass of Tennis Ball 1 Mass of Tennis Ball 1 Mass of Tennis Ball 1 Mass of Tennis Ball 1 10:47PM Wed May 15 X Radius of Tennis Ball 1 Radius of Tennis Ball 2 Radius of Tennis Ball 3 Radius of Tennis Ball 4 0.033932 0.034298 0.033502 0.034457 Dynamic equations: mg - D = ma mg 5 PCa Av' = ma a =g - kv2 k = 1 2m - pCa A To find the total time t as a function of x for the given system of differential equations: 1. du = 9 - kv2 dt 2. dx at = v we can follow these steps: Step 1: Solve at du = g - kv2 The equation = g - kv is separable. We can rewrite it as: 9 - kv2 = dt Integrate both sides: du g - kv2 dt Let's solve the integral on the left-hand side. We can use the partial fraction decomposition method. Let a = k, du g (1 - Q202 ) du 1 - a 2 2 = / at The integral J 1 0272 is a standard form that can be solved using the inverse hyperbolic tangent function:du 1 - Q212 = - tanh '(av) a Thus, - . tanh ' (av ) = t + C a Since v(0) = 0, we can find the constant C: tanh '(0) = 0 = C =0 ga Therefore, - tanh '(av) = t ga Solving for v, tanh (av) = gat av = tanh(gat) tanh(gat) U = a Step 2: Substitute v into do = v dx tanh(gat) dt a Step 3: Solve do _ tanh(gat) Separate variables and integrate:\fx = 1 tanh(u)du 1 x = 902 In(cosh(u)) + C Substituting back u = gat, x = go2 In(cosh(gat)) + C Since x (0) = 0. 0 = 1 go2 In(cosh (0)) + C = C =0 Thus, x = 902 In(cosh(gat)) Replacing a = 1 - In(cosh( Vgkt)) Solving for t as a function of x, kx = In(cosh(Vgkt) ) cosh(Vgkt) = ekx Vgkt = cosh -1(ekx)cosh(Vgkt) = er Vgkt = cosh -1(ekx) Therefore, the total time t as a function of x is: t(ac) = - 1 Vgk cosh (ekx) CO