Question

4. The vertical motion of mass A is defined by the relation x = 10 sin 2t + 15cos2t + 100, where x and t are expressed in mm and seconds, respectively. Determine (a) the position, velocity and acceleration of A when t = 1 s, (b) the maximum velocity and acceleration of A.

Answer 1

attached for your reference.

Answer 2

Answer:

(a) At t = 1:

Position is 115.34 mm

Velocity is 18.941 mm/sec

Acceleration is 61.36mm/sec² (retardation)

(b) Maximum velocity = 36.054 mm/sec

Maximum acceleration = 81.352 mm/sec²(retardation)

Explanation:

Given  x = 10 sin 2t + 15 cos 2t + 100                 --(i)

We know that velocity = $\frac{Displacement}{Time}$

=> V = $\frac{dx}{dt\\}$

Therefore, differentiating equation (i) with respect to t

=> V = $\frac{dx}{dt\\}$  = 10 cos 2t * 2 + 15 (- sin 2t) * 2 + 0

                = 20 cos 2t - 30 sin 2t                                     --(ii)

We know that Acceleration = $\frac{Velocity}{Time}$

=> A = $\frac{dV}{dt\\}$

Therefore, differentiating equation (ii) with respect to t

=>  A = $\frac{dV}{dt\\}$  = 20 (- sin 2t * 2) - 30 cos 2t * 2

                = -40 sin2t - 60 cos 2t                           --(iii)

(a) We need to find the position, velocity and acceleration when t = 1

Therefore, substituting t = 1 in equation (i) , (ii) and (iii)

=> Position =  10 sin 2(1) + 15 cos 2(1) + 100

                  = 10 * 0.0349 + 15* 0.9994 + 100

                  = 0.349 + 14.991 + 100

                  = 115.34 mm

Velocity = 20 cos 2(1) - 30 sin 2(1)  

             = 20 * 0.9994 - 30 * 0.0349

             = 19.988 - 1.047

             = 18.941 mm/sec

Acceleration = -40 sin2(1) - 60 cos 2(1)

                     = - 40 * 0.0349 - 60 * 0.9994

                     = -1.396 - 59.964

                     = -61.36mm/sec²

(b) To find maximum velocity, we find $\frac{d^2V}{dt^2}$

V = 20 cos 2t - 30 sin 2t    

=> $\frac{dV}{dt\\}$  = -40 sin2t - 60 cos 2t    

=> Differentiating above equation with respect to t

=> $\frac{d^2V}{dt^2}$ = -40 (cos 2t) * 2 - 60 (-sin 2t) * 2

          = - 80 cos 2t + 120 sin 2t                            --(iv)

Put  $\frac{dV}{dt\\}$  = 0

=> -40 sin2t - 60 cos 2t = 0

=> - 40 sin 2t = 60 cos 2t

=> tan 2t = -3/2

=> 2t = -56.31            

Substituting this in equation (iv)

=>  $\frac{d^2V}{dt^2}$ = - 80 cos (-56.31) + 120 sin (-56.31)

           = -80 * (0.5547) + 120 * (-0.832)

           = - 44.376 - 99.84

           = - 144.216 < 0      Therefore, maxima at 2t = -56.31

To find maximum velocity, substituting this value in equation (ii)

=> V =  20 cos (-56.31) - 30 sin (-56.31)

       = 20 * (0.5547) - 30 * (-0.832)

       = 11.094 + 24.96

       = 36.054 mm/sec

To find maximum Acceleration, we find $\frac{d^2A}{dt^2}$

A = -40 sin2t - 60 cos 2t  

=> $\frac{dA}{dt\\}$  = - 40 (cos 2t) * 2 - 60(-sin 2t) * 2

          = -80 cos 2t + 120 sin 2t

=> $\frac{d^2A}{dt^2}$ = -80 (-sin 2t) * 2 + 120 (cos 2t) * 2

          = 160 sin 2t + 240 cos 2t                            --(v)

Put  $\frac{dA}{dt\\}$  = 0

=> -80 cos 2t + 120 sin 2t = 0

=> 120 sin 2t = 80 cos 2t

=> tan 2t = 2/3

=> 2t = 33.69            

Substituting this in equation (v)

=>  $\frac{d^2A}{dt^2}$ = 160 sin (33.69) + 240 cos (33.69)

           = 160 * (0.5546) + 240 * (0.832)

           = 88.736 + 199.68

           = 288.416 > 0      (maxima at 2t 33.69 as acceleration = '-' ve )

To find maximum acceleration, substituting this value in equation (iii)

=> A = -40 sin (33.69) - 60 cos (33.69)

       = -40 * (0.5546) - 60 * (0.832)

       = - 22.184 - 59.168

       = -81.352 mm/sec²