Question

obtain an explicit formula for v(t)

dvb The integral on the left looks like an inverse trigonometric integral, but the sign is off. You look it up and obtain the formula U2-a2 du = a-tanh-1 (u/a) where tanh is the hyperbolic tangent function given by tanh(2) (e2 1)/e 1) and tanh 1 is its inverse function. Plug in the formula for this integral and apply tanh on both sides to obtain an explicit formula for v(t).

Answer 1

dV/dt = b/M (-Mg/b + v^2) = > integral 1/v^2 - Mg/b middot dv = integral b/M dt = b/M t + C = > integral 1/v^2 - a^2 middot dv = b/M t + C (1) where a = Mg/b Given that, integral 1/v^2 - a^2 dv = -1/a tan h^-1 (v/a) Hence (1) becomes: -1/a tan h^-1 (v/a) = b/M t + C = > tan h^-1 (v/a) = -ab/M t - ac = > v/a = tan h (-ab/M t - ac) = > v = a tan h (-ab/M t - ac) = > v = Mg/b tan h (-Mg/b middot b/M t - Mg/b c) = > V = Mg/b tan h (-gt - Mgc/b)