X= (7.1 m) cos[(5πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. at t = 1.6 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? also, what are the (e) frequency and (f) period of the motion?

Rather than answering your specific question I believe it's better to explain simple harmonic motion. We say some dynamical variable [tex]x[tex] is in simple harmonic motion if its value at time t is [tex]x(t)=A\cos(\omega t+\phi)[tex] for some constants [tex]A[tex] and [tex]\phi[tex]. The rate of change of [tex]x[tex] (in your case this is the velocity) as a function of time can be found by taking the derivative. In this case, [tex]v(t)=x'(t)=-A\omega\sin(\omega t + \phi)[tex]. The rate of change of the rate of change (in your case acceleration) can once again be found by taking the derivative [tex]a(t)=v'(t)=-A\omega^2\cos(\omega t +\phi)[tex]. The phase of the motion is just the argument of the cosine function, that is, [tex]\omega t+\phi[tex]. The frequency is the number of cycles which the motion makes in 1 second. This is of course just the inverse of the time it takes the oscillator in do one cycle. Since the cosine has period [tex]2\pi[tex] the time it take to make one cycle is [tex] T = \frac{2\pi}{\omega}[tex]. Therefore, the frequency is [tex]f=\frac{\omega}{2\pi}[tex].

In your problem [tex] A=7.1m[tex], [tex]\omega=5\pi rad/s[tex] and [tex]\phi=\pi/5rad[tex]. You've just gotta replace this values in the equations above to solve for what the problem asks.