Question

NIL

Answer 1

(a) Centripetal: $8.47 m/s^2$, tangential: $1.35 m/s^2$

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

$\alpha = 4.5\cdot 10^{-3}rad/s^2$

We can find the angular velocity of the car after half of a lap using the equivalent of the SUVAT equation for rotational motions:

$\omega_f^2 - \omega_i^2 = 2\alpha \theta$

where

$\omega_i = 0$ since the car starts from rest

$\theta = \pi$ is the angular displacement after half a lap

Solving for $\omega_f$,

$\omega_f = \sqrt{2\alpha \theta}=\sqrt{2(4.5\cdot 10^{-3})(\pi)}=0.168 rad/s$

Now we can find the centripetal acceleration with the formula:

$a_c = \omega^2 r = (0.168)^2 (300)=8.47 m/s^2$

while the tangential acceleration is simply given by

$a_t = \alpha r = (4.5\cdot 10^{-3})(300)=1.35 m/s^2$

(b) The mass of the car is not given, so it is not possible to find the forces.