Question

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 direction 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?

Answer 1

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Step-by-step 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=(a_t)/(t)\\\Rightarrow \alpha=(-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=(\omega_f^2-\omega_i^2^2)/(2\alpha)\\\Rightarrow \theta=(0^2-95^2)/(2* -24.28571)\\\Rightarrow \theta=185.80885\ rad=185.80885* (1)/(2\pi)\\\Rightarrow \theta=29.57239\ rev`

The number of revolutions is 29.57239

`\omega_f=\omega_i+\alpha t\\\Rightarrow t=(\omega_f-\omega_i)/(\alpha)\\\Rightarrow t=(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* 185.80885\\\Rightarrow s=52.026478\ m`

The distance the car travels is 52.026478 m