A spring is vibrating horizontally on a smooth level surface. The end of spring's distance from the equilibrium point by x(t) = 5 cos(t) where t is in seconds and x is in feet.

a) How far does the end of the spring start from the equilibrium point?

b) Find the velocity of the end of the spring when t = 2.

Question

A spring is vibrating horizontally on a smooth level surface. The end of spring's distance from the equilibrium point by x(t) = 5 cos(t) where t is in seconds and x is in feet.

a) How far does the end of the spring start from the equilibrium point?

b) Find the velocity of the end of the spring when t = 2.

anonymous · Answer

equilibrium position = 0

anonymous · Answer

For (a) all you it is asking you is for the end of the spring's distance from the equilibrium when t = 0. 
And for (b) you should remember that the derivative of direction is velocity so x'(t) = v(t). You can then find the derivative of 5cos(t) and solve for the value of the derivative at t = 2 for the velocity.