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A horse on the merry-go-round moves according to the equations r = 8 ft, u = (0.6t) rad, and z = (1.5 sin u) ft, where t is in seconds. Determine the cylindrical components of the velocity and acceleration of the horse when t = 4 s.

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hafsaabdulhai

Answer:

velocity = 0.664 ft/s, acceleration= 0.365 ft/s²

Explanation:

z= 1.5 sin 0.6t

angular velocity= dz/dt= d(1.5sin0.6t)/dt

                       = 0.9cos0.6t

                    at t=4

                       =0.9cos(0.6×4)

                        = -0.664 ft/s

Acceleration= dv/dt

                      = d(0.9cos0.6t)/dt

                      = -0.54sin0.6t

             at t=4

                   acceleration= -0.54sin(0.6××4)

                                           = 0.365 ft/s²

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