Answer:
A) About 13.83 m/s²
B) 30.25 m.
Explanation:
Part A)
Centripetal acceleration is given by:
[tex]\displaystyle a_c = \frac{v^2}{r}[/tex]
Hence, the centripetal acceleration of the car is:
[tex]\displaystyle a_c = \frac{(22\text{ m/s})^2}{35\text{ m}} \approx 13.83\text{ m/s$^2$}[/tex]
Part B)
The force of friction supplies the centripetal force:
[tex]\displaystyle F_f = \frac{mv^2}{r}[/tex]
Because the maximum friction that can be supplied is 20,000 N, we have that:
[tex]\displaystyle \frac{mv^2}{r} \leq 20000\text{ N}[/tex]
Therefore:
[tex]\displaystyle \begin{aligned} (1250\text{ kg})(22\text{ m/s})^2} & \leq 20000\text{ N}\cdot r \\ \\ r & \geq \frac{(1250\text{ kg})(22\text{ m/s})^2}{20000\text{ N}} \\ \\ r & \geq 30.25\text{ m} \end{aligned}[/tex]
The radius of the smallest circle the car can drive is hence 30.25 m.
