The frictional force between the tires and the road prevent the car from skidding off the road due to centripetal force.

If the frictional force is less than the centripetal force, the car will skid when it navigates a circular path.

The diagram below shows that when the car travels at tangential velocity, v, on a circular path with radius, r, the centripetal acceleration of v²/ r acts toward the center of the circle.

The resultant centripetal force is (mv²)/r, which should be balanced by the frictional force of μmg, where μ = coefficient of kinetic friction., and mg is the normal reaction on a car with mass, m.

This principle is applied on racing tracks, where the road is inclined away from the circle to give the car an extra restoring force to overcome the centripetal force.

8 0

2 years ago

3logbase5(x^2+9)-6=0

Pavlova-9 [17]

$3\log_5(x^2+9)-6=0$
$3\log_5(x^2+9)=6$
$\log_5(x^2+9)=2$
$5^{\log_5(x^2+9)}=5^2$
$x^2+9=25$
$x^2=16$
$x=\pm\sqrt{16}$
$x=\pm4$

5 0

3 years ago