angular motion with constant acceleration?

Question

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to theta(t)=theta_0+omega_0t+0.5alpha t^2. At time t=t_1, particle B, which also undergoes constant angular accelaration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time t=0.
How long after the time t_1 does the angular velocity of B have to be to equal A's?
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ok, here we go:
i know that i have to write expressions for the angular velocity of A and B as functions of time, and solve for t-t1???
For A, omega(t) = omega_0 + alpha(t) and
for B, omega(t) = 0.5 omega_0 + 2alpha(t-t_1)
so when i equate them and solve for t-t_1, i get something that is not the right answer!

please help :(

Answer 1

Let t = the time after t1 when the angular velocities are equal. Then
ω0 + α(t + t1) = (1/2)ω0 + 2αt
2ω0 + 2αt + 2αt1 = ω0 + 4αt
2αt = ω0 + 2αt1
t = t1 + ω0/2α

Answer 2

we are asked approximately acceleration yet we don't be attentive to the time the two, so we prefer 2 self sustaining equations to remedy this subject, a million & 2 will do. a million) th = Th0 + W0 * t + A * t^2 / 2 2) w = W0 + A * t the place; th = 5rev, Th0 = 0, W0 = 33 a million/3rpm, w = 45rpm remedy #2 for t t = (w - W0) / A substitute in a million th = Th0 + W0 * (w - W0) / A + A * ((w - W0) / A)^2 / 2 = Th0 + W0 * w / A - W0^2 / A + (w^2 - 2 * w * W0 + W0^2) / (A * 2) = Th0 - W0^2 / 2A + w^2 / 2A = Th0 + (w^2 - W0^2) / 2A It looks like we've won the equation selection 3 ;) remedy for A A = (w^2 - W0^2) / 2th = ((45rpm)^2 - (33.33rpm)^2) / (2 * 5r) = ninety one.4rev according to minute^2