Respuesta :
Answer:
A) Angular acceleration = -2.47 rad/s²
B) 23.54 seconds
C) The total distance covered = 294.23m
Explanations:The number of revolutions = 62
Angular distance, θ = 62 x 2π
θ = 62 x 2 x 3.142
θ = 389.608 radians
Diameter, d = 0.86 m
Radius, r = d/2 = 0.86/2
r = 0.43m
Initial velocity, v₁ = 90 km/h = 90 x (1000/3600)
v₁ = 25 m/s
Angular velocity, w₁ = v₁ / r
w₁ = 25/0.43
w₁ = 58.14 rad/s
Final velocity, v₂ = 59 km/h = 59 x (1000/3600)
v₂ = 16.39 m/s
Angular velocity, w₂ = v₂ / r
w₂ = 16.39 / 0.43
w₂ = 38.12 rad/s
Using the equation of motion:
[tex]\begin{gathered} w^2_2=w^2_1\text{ + 2}\alpha\theta \\ 38.12^2=58.14^2\text{ + 2}\alpha(389.608) \\ 38.12^2-58.14^2=\text{ }779.216\alpha \\ 779.216\alpha\text{ = }-1927.1252 \\ \alpha\text{ = }\frac{-1927.1252}{779.216} \\ \alpha\text{ = }-2.47rad/s^2 \end{gathered}[/tex]Angular acceleration = -2.47 rad/s²
B) Amount of time required for the car to stop if it continues to decelerate at this rate
Initial angular speed, w₁ = 58.14 rad/s
When the car stops, final angular speed, w₂ = 0 rad/s
Using the equation of motion below:
[tex]\begin{gathered} w_2=w_1+\text{ }\alpha t \\ 0\text{ = 58.14 + (-2.47)t} \\ -2.47t\text{ = -58.14} \\ t\text{ = }\frac{-58.14}{-2.47} \\ t\text{ = }23.54\text{ seconds} \end{gathered}[/tex]C) The total distance
Use the equation of motion below:
[tex]\begin{gathered} S=v_1\text{t + }\frac{1}{2}at^2 \\ a\text{ = }\alpha r \\ a\text{ = (-2.47)(0.43)} \\ a\text{ = }-1.0621m/s^2 \end{gathered}[/tex][tex]\begin{gathered} S=v_1\text{t + }\frac{1}{2}at^2 \\ S\text{ = }25(23.54)+0.5(-1.0621)(23.54)^2 \\ S\text{ = }588.5-294.27 \\ S\text{ = }294.23\text{ m} \end{gathered}[/tex]The total distance covered = 294.23m
