No wonder you are having problems, your equation for SHO is messed up.

y = A cos(omega*t + phi)

where omega is the (angular) frequency and phi is the (angular) displacement. (Note: your book might use a different terminology for phi.)

For all three of these we have the same frequency. So, the period T is 10 s. Thus omega = 2*pi/T = 0.62832 Hz (technically the unit is rad/s.)

We also know that A = 3.5 ft because A is the magnitude of the max or min value of the wave.

a) y(0) = 3.5 ft.

y(t) = 3.5*cos(0.62832*t + phi)

At t = 0 s:

y(0) = 3.5*cos(0.62832*0 + phi) = 3.5

Thus cos(phi) = 1 ==> phi = 0 rad.

So y(t) = 3.5*cos(0.62832*t)

b) v = dy/dt = -3.5*0.62832*sin(0.62832*t)

v(t) = -2.1991*sin(0.62832*t) (ft/s)

c) The velocity is greatest when dv/dt = 0, so:

dv/dt = -2.1991*0.62832*cos(0.62832*t)

0 = 7.69690*cos(0.62832*t) ==> cos(0.62832*t) = 0

Thus

0.62832*t = pi/2 + n*pi where n = 0, 1, 2, 3,....

0.62832*t = pi/2 + n*pi ==> t = 2.5 + 5n (s)

The velocity at these times are:

v(max) = -2.1991*sin(0.62832*[2.5 + 5n])

v(max) = -2.1991*sin(pi/2 + n*pi) <-- These are the exact values. The approximated values work nearly as well.

Now, sin(A + B) = sin(A)cos(B) + sin(B)cos(A)

So sin(pi/2 + n*pi) = sin(pi/2)cos(n*pi) + sin(n*pi)cos(pi/2)

sin(pi/2) = 1

cos(n*pi) = (-1)^{n}

sin(n*pi) = 0 for all n

cos(pi/2) = 0

Thus sin(pi/2 + n*pi) = (-1)^{n}

Finally:

v(max) = -2.1991*(-1)^{n} = 2.1991*(-1)^{n+1} (ft/s)

Probably your book was REALLY asking for the speed at these times which is a scalar, rather than the velocity which is a vector. The speed at these times is the magnitude of the velocity, and thus will be

2.1991 ft/s.

-Dan