Question

A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction between the tires and the road is μ = 0.800. If the car maintains a constant speed of 28 m/s, it will A. first veer toward the center for the first quarter-circle, then veer toward the outside for the second quarter circle. B. tend to veer toward the outside of the circle.

Answer 1

As 28m/s = 28m/s

Step-by-step explanation:

r = the radius of the curve

m = the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

`a = (v^(2))/(r)`

therefore

`\mu mg = m (v^(2))/(r) \\\\ \mu = (v^(2))/(rg)`

`v = √(\mu rg)`

`\mu = √(0.8 * 100*9.8) \\\\= 28m/s`

As 28m/s = 28m/s