Tangential and centripetal acceleration

[RaamGeneral]

Hello.

Sorry for this very silly question, but I struggled myself too long on this (it was on the test and fortunataly I could answer correctly, but have not understood how to solve that).

I know a particle, in a certain instant, has position $(x_0,y_0,z_0)$
I know the velocity magnitude $\vec{v} = v_p \hat{\tau}$

where tau is the tangent unit vector.
And I know also the acceleration vector:
$\vec{a}=a_0\hat{x}+a_0\hat{y}$I have to find the tangential and centripetal acceleration of the particle at that instant.Thanks.

 

[td21]

tangential acc is a times tau.
you can then figure out centripetal acceleration.

 

[RaamGeneral]

Ok, I think I should insert the numbers here, because I have to figure out the numerical value and I do know nothing about tau.
position $(x_0 = \sqrt 3 m , y_0=1 m , z_0 =2.12 m)$
$v_p = 4.78 m/s$
$a_0 = 6.01 m/s^2$

I have only this, nothing less, nothing more.

 

[RaamGeneral]

Please, it's my last unfinished business.
This is the original text. It's in italian, but mathematics is a universal language.

 

[td21]

ep = cos30 ex + sin30 ey
then take the dot product of ep and a
i.e.
(cos30 ex + sin30 ey)⋅(6.01 ex + 6.01 ey)