Question

A car of mass 1250 kg drives around a curve with a speed of 22 m/s.

a. if the radius of the curve is 35 m, what is the centripetal acceleration of the car? (3 points)
b. the maximum friction that the car's tires can apply is 20,000 n. what is the radius of the smallest circle the car can drive in at a speed of 22 m/s? (3 points)

Answer 1

A) About 13.83 m/s²

B) 30.25 m.

Step-by-step explanation:

Part A)

Centripetal acceleration is given by:

`\displaystyle a_c = (v^2)/(r)`

Hence, the centripetal acceleration of the car is:

`\displaystyle a_c = \frac{(22\text{ m/s})^2}{35\text{ m}} \approx 13.83\text{ m/s$^2$}`

Part B)

The force of friction supplies the centripetal force:

`\displaystyle F_f = (mv^2)/(r)`

Because the maximum friction that can be supplied is 20,000 N, we have that:

`\displaystyle (mv^2)/(r) \leq 20000\text{ N}`

Therefore:

`\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}`

The radius of the smallest circle the car can drive is hence 30.25 m.