A package is pushed up an incline at x = 0 with an initial speed v0. The incline is coated with a thin viscous layer so that the acceleration of the package is given by a = -(g sin(theta) + n v), where g is the acceleration due to gravity, n is a constant, and v is the velocity of the package. If theta = 25 A package is pushed up an incline at x = 0 with an initial speed v0. The incline is coated with a thin viscous layer so that the acceleration of the package is given by a = -(g sin(theta) + n v), where g is the acceleration due to gravity, n is a constant, and v is the velocity of the package. If theta = 25degree, v0 = 10 m/s, and n = 9 s^-1, determine the distance d traveled by the package before it comes to a stop. Solution dv/dt = a = -(g*sin(theta)+n*v) Integrating with respect to t, then dv/(gsin(25)+9v) = dt then ln(4.15+9v) = -t + k        (k is constant) at t=0, v=10m/s, then k = 4.54 => ln(4.15 + 9v) = -t + 4.54 v = -.46 + 10.46e^-t = dx/dt again if we integrate, x = -.46t - 10.46e^-t + m (where m is constant) at t=0, x=0, thenn m = 10.92 then x = -.46t - 10.46*e^-t + 10.92 at t=3.12 sec, v = 0, Distance (x) = 9.02 m .