Question

An car traveling at 110.0 km/h has tires of 67.0 cm diameter. If the car is brought to a stop uniformly in 45.0 complete turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels?

Answer #1

u (speed) = 110km/hr

= 110 * 5/18 m/s

= 30.55 m/s

Diameter = 67 cm

Radius = 67/2 = 33.5 cm

= 0.335 m

No. of turns (N) = 45

V ( final speed) = 0 m/s (Rest)

Using, equation of motion

w^2 - W_o^2 = 2

where w is the final angular velocity

w_o = initial angular velocity

is angular acceleration

angular displacement

we know that, V = R w

Substituting the given values in the above formula

0 - (30.55/0.335)^2 = 2** (45*2)

-8316.35 = 565.58

= - 14.7 rad/sec^2

Magnitude of angular acceleration of the wheels = **14.7
rad/sec^2**

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An car traveling at 110.0 km/h has tires of 67.0 cm diameter. If
the car is brought to a stop uniformly in 45.0 complete turns of
the tires (without skidding), what is the magnitude of the angular
acceleration of the wheels?
A.
14.7 rad/s2
B.
9.7 rad/s2
C.
10.7 rad/s2
D.
-14.7 rad/s2

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has wheels of 30-in diameter.
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If the wheels are brought to a stop uniformly in 30 turns, what
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How far does the car advance during this braking period?

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B)If the car continues to decelerate at this rate, how much more
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