Answer:
18.6 m/s
Explanation:
The frictional force acting on the car is given by:
[tex]F_f = \mu m g=(0.40 )(1200 kg)(9.81 m/s^2)=4709 N[/tex]
But the frictional force also corresponds to the centripetal force that keeps the car in circular motion in the turn:
[tex]F_f = F_c = m \frac{v^2}{r}[/tex]
Where v is the maximum speed the car can achieve remaining in the turn. Substituting r=88.0 m and re-arranging the formula, we can find the value of v:
[tex]v=\sqrt{\frac{Fr}{m}}=\sqrt{\frac{(4709 N)(88.0 m)}{1200 kg}}=18.6 m/s[/tex]
