Vector Problem In Polar Coordinates

A particle moves in polar coordinates with r=20t^3 and theta=5t. Find (vector)v (for velocity) and (vector)a (for acceleration) as functions of time. I deriviated r and theta and got the following:
r = 20t^3
r' = 60t^2
r'' = 120t

theta = 5t
theta' = 5
theta'' = 0

For v I got = (60t^2)r((^)(hat)) +(60t^3)(theta hat)

For a I got (r hat)(120t - 500t^3)+(theta hat)(600t^2)

Can you please check my work to make sure that I got the correct answers.
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You can do this (or check your work) in cartesian coordinates.

(r = 20t^3, θ = 5t) = ( x = 20t^3 cos 5t, y = 20t^3 sin 5t )
(d/dt) (r = 20t^3, θ = 5t) = (d/dt) ( x = 20t^3 cos 5t, y = 20t^3 sin 5t ) = ( x = 60t^2 cos 5t - 100 t^3 sin 5t, y = 60t^2 sin 5t + 100t^3 cos 5t )
(d^2/dt^2) (r = 20t^3, θ = 5t) = (d/dt) ( x = 60t^2 cos 5t - 100 t^3 sin 5t, y = 60t^2 sin 5t + 100t^3 cos 5t ) = ( x = 120 t cos 5t - 600t^2 sin 5t - 500 t^3 sin 5t, y = 120 t sin 5t + 600 t^2 cos 5t - 500 t^3 sin 5t)


Alternately, you can work entirely in polar coordinates. But the formulas are complicated. And I think you have the wrong expressions.

http://themcclungs.net/physics/download/H/...Coordinates.pdf
http://www.ecourses.ou.edu/cgi-bin/eBook.c...1.6&page=theory

Where do you think I went wrong. Are both my velocity and acceleration equations incorrect?

r = 20 t^3, r' = 60 t^2, r'' = 120 t
θ = 5t , θ' = 5, θ'' = 0

(So far, so good)

From θ we get our unit basis vectors:
r^ = cos(5t)x̂ + sin(5t)ŷ
θ^ = -sin(5t)x̂ + cos(5t)ŷ

r = r r^ = 20t^3 r^
(d/dt)r = r' r^ + r θ' θ^ = 60 t^2 r^ + 100 t^3 θ^
(d^2/dt^2)r = (r'' - r θ'^2) r^ + (2 r' θ' + rθ'') θ^ = ( 120 t - 500 t^3) r^ + (600 t^2 + 0) θ^

So you got acceleration right and velocity half-right.

Thank you I saw where I messed up at (20t^3)(5)(theta hat). I I solved this part of the equation I forgot to multiply the 5 times 20 and for some reason I put 60. Just a stupid mistake. Thank you for the help!!