During a very quick stop, a car decelerates at 28.4 rad/s? . assume the tires initially rotated in the positive direction and radius of tires is 0.32m a) how many revolutions do the tires make before coming to rest, given their initial angular velocity is 93.0 rad/s? b) how long does the car take to stop completely? c) what was the car's initial velocity (in m/s)? d) how far the car moves forward before it stops?

a) 24

b) 3.3 sec

c) 29.8 m/s

d) 48.85 m

Explanation:

a)

α = angular acceleration = - 28.4 rad/s²

r = radius of the tire = 0.32 m

w₀ = initial angular velocity = 93 rad/s

w = final angular velocity = 0 rad/s

θ = angular displacement

Using the equation

w² = w₀² + 2αθ

0² = 93² + 2 (- 28.4) θ

θ = 152.3 rad

n = number of revolutions

Number of revolutions are given as

b)

t = time taken to stop

using the equation

w = w₀ + αt

0 = 93 + (- 28.4) t

t = 3.3 sec

c)

v₀ = initial velocity of the car

initial velocity of the car is given as

v₀ = r w₀ = (0.32) (93) = 29.8 m/s

d)

v = final velocity = 0 m/s

a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²

d = distance traveled by car before stopping

Using the equation

v² = v₀² + 2 a d

0² = 29.8² + 2 (- 9.09) d

d = 48.85 m

pe = weightxheight = 2000x1000=2,000,000joules

i think its c if i'm wrong