Answer:
-25 rad/s²
29 times
3.8 seconds
50.54 m
26.6 m/s
Explanation:
[tex]\omega_f[/tex] = Final angular velocity
[tex]\omega_i[/tex] = Initial angular velocity
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
a = Acceleration = -7 m/s² (negative because of deceleration)
r = Radius of wheel = 0.28 m
Angular acceleration is given by
[tex]\alpha=\frac{a}{r}\\\Rightarrow \alpha=\frac{-7}{0.28}\\\Rightarrow \alpha=-25\ rad/s^[/tex]
Angular acceleration of the wheel is -25 rad/s²
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-25}\\\Rightarrow t=3.8\ s[/tex]
It took 3.8 seconds for the car to stop
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=95\times 3.8+\frac{1}{2}\times -25\times 3.8^2\\\Rightarrow \theta=180.5\ rad[/tex]
[tex]\frac{180.5}{2\pi}=28.72746\ rev[/tex]
The wheel rotated 29 times.
The
Initial velocity
[tex]u=r\omega_i\\\Rightarrow u=0.28\times 95\\\Rightarrow u=26.6\ m/s[/tex]
The car’s initial velocity is 26.6 m/s
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=26.6\times 3.8+\frac{1}{2}\times -7\times 3.8^2\\\Rightarrow s=50.54\ m[/tex]
Distance covered in the time is 50.54 m
If the distance is found in meters we get 50.54 m which is very low. Considering the initial velocity is 95.76 km/h. Coming to a stop in that distance is very low. So, the values do not seem reasonable. At least a 100 m is required to stop from that speed.
