a = (0-50) / 5 = -10m/s^2.
C = pi*D = 3.14 * (2*0.25) = 1.57m/rev.
V=10m/S^2 * (1/1.57)rev/m * 6.28rad/rev
= 40rad/s^2.
C = pi*D = 3.14 * (2*0.25) = 1.57m/rev.
V=10m/S^2 * (1/1.57)rev/m * 6.28rad/rev
= 40rad/s^2.
α = ωf - ωi / t
where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time taken.
To compute the angular acceleration, we need to calculate the initial and final angular velocities.
The linear velocity of the car is given as 50 m/s. The linear velocity of a point on the edge of a rotating object is related to the angular velocity by the equation:
v = r * ω
where v is the linear velocity, r is the radius, and ω is the angular velocity.
Rearranging the equation to solve for ω, we have:
ω = v / r
Plugging in the values, we get:
ω = 50 m/s / 0.25 m
ω = 200 rad/s
Therefore, the initial angular velocity of the car's tires is 200 rad/s.
The final angular velocity is zero since the car comes to a stop. So, ωf = 0 rad/s.
Substituting the values in the formula for angular acceleration, we get:
α = 0 - 200 rad/s / 5 s
α = -40 rad/s^2
Hence, the angular acceleration of the car's tires is -40 rad/s^2. The negative sign indicates that the tires are decelerating or slowing down.