Answer:
1.18 m/s²
Explanation:
The linear speed = 13.0s
[tex]v= v_i + at\\v = 0 + 0.42(13)\\v = 5.46m/s[/tex]
where v(i) is the initial linear speed
the radial acceleration of the car afte (t) = 13s is
[tex]a_r = \frac{v^2}{r} \\= \frac{5.46^{2} }{27} \\= 1.104m/s^2[/tex]
the magnitude of the net acceleration is
[tex]a_net = \sqrt{1.104^2 + 0.1764^2} \\= 1.18m/s^2[/tex]
