A spaceship negotiates a circular turn of radius......?

Question

A spaceship negotiates a circular turn of radius 2885 km at a speed of 31000 km/h.
(a) What is the magnitude of the angular velocity?
in rad/s
(b) What is the magnitude of the radial acceleration?
in m/s2
(c) What is the magnitude of the tangential acceleration?
in m/s2

i'm stumped with this hw problem, i know that C)=0, but im clueless on A and B.

Answer 1

a) assuming the speed given is tangential the angular velocity
w=V/R
w=31000 km/h /2885 km =10.75 rad/h or
w=2.99 e-3 rad/sec
b) ar=V^2/R`=(31,000,000)^2 / 2,885,000 =25.7.m/sec^2
or ar=w^2 R=(2.99 e-3 rad/sec) ^2 x 2,885,000 =25.7m/s Check
c) at=0 since the tangential speed is constant
at=change of velocity with respect to time.