Position of an oscillating object

[ChloeYip]

Homework Statement

The position of an object that is oscillating on an ideal spring is given by the equation x =
(12.3 cm) cos[(1.26s-1)t]. At time t = 0.815 s,
(a) how fast is the object moving?
(b) what is the magnitude of the acceleration of the object?

Homework Equations

As follow

The Attempt at a Solution

dx/dt = -12.3*(1.26)sin(1.26*.815) = -0.2777 cm/s but the answer is -13.3cm/sec
dx^2/dt^2= -12.3*(1.26)^2*cos(1.26*.815) = -19.524 cm/s^2 but the answer is -10.1cm/sec^2

I have only little time before test.hope there is someone help me soon.tell me what's wrong with it.
(Please don't ask me to guess... i m not good at that and i really have not time left...)
Thank you very much for helping me

 

[vela]

$12.3 \ne 1.23$ for one thing.

 

[ChloeYip]

Thanks for reminding
Thats only multiple of 10, still canet get the answer
Also, the calculated answer is not affect by this typo
Sorry for typo

 

[vela]

Make sure your calculator is in radian mode.

 

[ChloeYip]

Ahhh yes
Thanksssssss