Answer:
(a) 1.64 rad/s²
(b) 39.94 rad
Step-by-step explanation:
(a) From the question,
α = a/r................ Equation 1
Where a = linear acceleration if the motorcycle wheel, r = radius of the wheel, α = angular velocity of the wheel.
But,
a = (v-u)/t.............. Equation 2
Given: v = 1.4 m/s, u = 7.8 m/s, t = 5.8 s
Substitute into equation 2.
a = (1.4-7.8)/5.8
a = -1.1 m/s².
Note: The negative sign shows that the motorcycle is decelerating
Also given: r = 0.67 m.
Substitute into equation 1
α = 1.1/0.67
α = 1.64 rad/s²
(b)
Θ = s/r................... Equation 3
Where s = linear displacement, Θ = angular displacement.
But,
s = (v²-u²)/2a.......... Equation 2
s = (1.4²-7.8²)/[2×(-1.1)]
s = -58.88/-2.2
s = 26.76 m.
therefore,
Θ = 26.76/0.67
Θ = 39.94 rad.