Answer:
The magnitude of net linear acceleration of the car after 16 s later is [tex]1.72\ m/s^2[/tex].
Explanation:
It is given that,
Initial speed of the car, u = 0 (at rest)
Radius of the circular track, r = 34 m
Acceleration of the car, [tex]a=0.47\ m/s^2[/tex]
We need to find the magnitude of its net linear acceleration 16.0 s later. It is equal to the resultant of radial and linear acceleration. The linear speed of the car after 15 seconds. So,
[tex]v=u+at\\\\v=0+at\\\\v=0.47\times 16\\\\v=7.52\ m/s[/tex]
The radial acceleration of the car is given by :
[tex]a'=\dfrac{v^2}{r}\\\\a'=\dfrac{(7.52)^2}{34}\\\\a'=1.66\ m/s^2[/tex]
So, net acceleration of the car is given by :
[tex]a_n=\sqrt{a^2+a'^2} \\\\a_n=\sqrt{(0.47)^2+(1.66)^2} \\\\a_n=1.72\ m/s^2[/tex]
So, the magnitude of net linear acceleration of the car after 16 s later is [tex]1.72\ m/s^2[/tex]. Hence, this is the required solution.
