Answer
(a) $\omega(t) = (6.40~rad/s^2)~t - (1.50~rad/s^3)~t^2$
(b) $\alpha(t) = (6.40~rad/s^2) - (3.00~rad/s^3)~t$
(c) The maximum positive angular velocity is 6.83 rad/s.
$t = 2.13~s$
Work Step by Step
(a) $\theta(t) = (3.20~rad/s^2)~t^2 - (0.500~rad/s^3)~t^3$
$\omega(t) = \frac{d\theta}{dt}$
$\omega(t) = (6.40~rad/s^2)~t - (1.50~rad/s^3)~t^2$
(b) $\omega(t) = (6.40~rad/s^2)~t - (1.50~rad/s^3)~t^2$
$\alpha(t) = \frac{d\omega}{dt}$
$\alpha(t) = (6.40~rad/s^2) - (3.00~rad/s^3)~t$
(c) The maximum positive angular velocity occurs when $\alpha = 0$.
$\alpha(t) = (6.40~rad/s^2) - (3.00~rad/s^3)~t = 0$
$(6.40~rad/s^2) = (3.00~rad/s^3)~t$
$t = \frac{6.40~rad/s^2}{3.00~rad/s^3}$
$t = 2.13~s$
We can find the maximum positive angular velocity.
$\omega(t) = (6.40~rad/s^2)~t - (1.50~rad/s^3)~t^2$
$\omega = (6.40~rad/s^2)(2.13~s)- (1.50~rad/s^3)(2.13~s)^2$
$\omega = 6.83~rad/s$
The maximum positive angular velocity is 6.83 rad/s.
