(a) Centripetal: [tex]8.47 m/s^2[/tex], tangential: [tex]1.35 m/s^2[/tex]
The radius of the track (which is the distance of the car from the centre of the circular path) is
r = 0.30 km = 300 m
The angular acceleration is
[tex]\alpha = 4.5\cdot 10^{-3}rad/s^2[/tex]
We can find the angular velocity of the car after half of a lap using the equivalent of the SUVAT equation for rotational motions:
[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta[/tex]
where
[tex]\omega_i = 0[/tex] since the car starts from rest
[tex]\theta = \pi[/tex] is the angular displacement after half a lap
Solving for [tex]\omega_f[/tex],
[tex]\omega_f = \sqrt{2\alpha \theta}=\sqrt{2(4.5\cdot 10^{-3})(\pi)}=0.168 rad/s[/tex]
Now we can find the centripetal acceleration with the formula:
[tex]a_c = \omega^2 r = (0.168)^2 (300)=8.47 m/s^2[/tex]
while the tangential acceleration is simply given by
[tex]a_t = \alpha r = (4.5\cdot 10^{-3})(300)=1.35 m/s^2[/tex]
(b) The mass of the car is not given, so it is not possible to find the forces.
