Question

A particle oscillates along the x-axis according to equation x=0.05sin(5t-180devided by6)where x is in metre and t is in second find its velocity at t=0second

Answer 1

dx/dt=0. 05 cos(5t-180)/6 ×5/6 , velocity = 0. 05 cos(-30)×5/6 =0. 05×5/6×Underroot3/2

Answer 2

Answer:

Hey mate

Here is your answer

For velocity

For velocityV = d x/dt = 0.05 cos(5t- pi/6) * 5

For velocityV = d x/dt = 0.05 cos(5t- pi/6) * 5Hence at t = 0

For velocityV = d x/dt = 0.05 cos(5t- pi/6) * 5Hence at t = 0V = 0.25 cos (-pi/6)

For velocityV = d x/dt = 0.05 cos(5t- pi/6) * 5Hence at t = 0V = 0.25 cos (-pi/6)V = 0.25 *(1/2) *\sqrt3

For velocityV = d x/dt = 0.05 cos(5t- pi/6) * 5Hence at t = 0V = 0.25 cos (-pi/6)V = 0.25 *(1/2) *\sqrt3V = 25/100 *(1/2) */sqrt3

For velocityV = d x/dt = 0.05 cos(5t- pi/6) * 5Hence at t = 0V = 0.25 cos (-pi/6)V = 0.25 *(1/2) *\sqrt3V = 25/100 *(1/2) */sqrt3V = 1/4 * 1/2 */sqrt3

For velocityV = d x/dt = 0.05 cos(5t- pi/6) * 5Hence at t = 0V = 0.25 cos (-pi/6)V = 0.25 *(1/2) *\sqrt3V = 25/100 *(1/2) */sqrt3V = 1/4 * 1/2 */sqrt3V = /sqrt3/8

HOPE THIS HELPS YOU

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