Answer:
The centripetal acceleration of the car is 50.4 m/s²
Explanation:
Given;
radius of the track, r = 0.3km = 300 m
constant angular acceleration, α = 4.5 x 10⁻³ rad/s²
during half of a lap, θ = π
Determine time of motion
θ = ωt + ¹/₂αt²
π = 0 + ¹/₂αt²
t² = 2π / α
t² = ( 2 x 3.142) / (4.5 x 10⁻³)
t² = 1396.44
t = √1396.44
t = 37.369 s
Determine the angular velocity
[tex]\omega_f = \omega_i + \alpha t\\\\\omega_f = 0 \ + \ 37.369*4.5*10^{-3}\\\\\omega_f = 0.168 \ rad/s[/tex]
Determine the centripetal acceleration of the car
[tex]\alpha _c = \omega^2 r\\\\\alpha _c = 0.168*300\\\\\alpha _c =50.4 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the car is 50.4 m/s²
