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alexandr1967 [171]
3 years ago
13

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

on for the rotation of the tires (and that they do not slip on the pavement).(a) What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement? (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95.0 rad/s ? (c) How long does the car take to stop completely? (d) What distance does the car travel in this time?
Physics
1 answer:
geniusboy [140]3 years ago
8 0

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

a_t = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

\omega_i = Initial angular velocity = 95 rad/s

\theta = Angle of rotation

\omega_f = Final angular velocity

t = Time taken

Angular acceleration is given by

\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2

The angular acceleration is -24.28571 rad/s²

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev

The number of revolutions is 29.57239

\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s

The time it takes for the car to stop is 3.91176 seconds

Linear distance

s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m

The distance the car travels is 52.026478 m

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