Question

A simple harmonic oscillator's velocity is given by 

v_y(t) = (0.930 m/s)sin(11.0t − 6.45).

 Find the oscillator's position, velocity, and acceleration at each of the following times. (Include the sign of the value in your answer.)

(a)    

t = 0

position	-0.0834  m
velocity	-0.154  m/s
acceleration	10.1  m/s2

(b)    

t = 0.500 s

position	-0.0492  m
velocity	-0.756  m/s
acceleration	5.95  m/s2

(c)    

t = 2.00 s

position	0.0835 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m
velocity	0.146 Your response differs from the correct answer by more than 100%. m/s
acceleration	-10.1 The response you submitted has the wrong sign. m/s2

I used,

y(t)= −0.0845 cos(11.0t − 6.45) m to solve position for 1a, 1b, and 1c 

ay(t) =10.2 cos(11.0t − 6.45) m/s^2 to solve velocity for 1a, 1b, and 1c, and

used vy(t) = (0.930 m/s)sin(11.0t − 6.45) to solve acceleration for 3a, 3b, and 3c.

As a result, 

I got a question, 1a, and 1b correct. But how come I got 1c completely wrong. What formula will help me to get the right answer? 

Can you provide me a solution for 1c with step by step and proper formula to get into the correct answer PLEASE!

Thanks.