The driver of a car traveling at 21.6 m/s applies the brakes and undergoes a constant deceleration of 1.86 m/s 2 . How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of 0.35 m

each tire made 57.03 revolutions before the car halted

Explanation:

Given that;

Initial Velocity u =  21.6 m/s

final velocity v = 0 m/s

acceleration = -  1.86 m/s²

radii r = 0.35 m

first we calculate distance travelled by the car to stop;

v² - u² = 2as

s = v² - u² / 2a

we substitute

s = (0² - (21.6)² / 2 × (-1.86)

s = (0 - 466.56) / ( - 3.72 )

s = 125.419 m

Now Distance traveled by each tire in 1 revolution = 2πr

i.e 2πr is travelled in 1 revolution

so

n×2πr = 125.419 m

n = 125.419 m / 2πr

n = 125.419 m / 2π(0.35)

n = 125.419 / 0.7π

n = 57.03 revolutions

Therefore, each tire made 57.03 revolutions before the car halted

568.98

explanation:

its math

c. 7/3

step-by-step explanation:

a rational number is like pi- it is a repeating decimal.

a. is -1.8

b. 2.236068

c. 2.3333333333333333333

d.   is 3

i hope that ! =)