A package is pushed up an incline at x = 0 with an initial speed v0. The incline

Question

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.

Explanation / Answer

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

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