To test the performance of its tires, a car
travels along a perfectly flat (no banking) cir-
cular track of radius 564 m. the car increases
its speed at uniform rate of
u du = 3.62 m/s?
dt
until the tires start to skid.
if the tires start to skid when the car reaches
a speed of 35 m/s, what is the coefficient of
static friction between the tires and the road?

0.43

Explanation:

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

There are friction forces in two directions: centripetal and tangential.  The centripetal acceleration is:

ac = v² / r

ac = (35 m/s)² / 564 m

ac = 2.17 m/s²

The total acceleration is:

a = √(ac² + at²)

a = √((2.17 m/s²)² + (3.62 m/s²)²)

a = 4.22 m/s²

Sum of forces:

∑F = ma

Nμ = ma

mgμ = ma

μ = a / g

μ = 4.22 m/s² / 9.8 m/s²

μ = 0.43

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

not moving. pos accel,