Answer
$0.6$ km/sec and $\dfrac{18}{\pi}rev/min $
Work Step by Step
Since, velocity $v=\dfrac{dx}{dt}$
Then $v=\sec^2 \theta\dfrac{d\theta}{dt}=(-0.6)\sec^2 \theta$
Plug $ \theta=0$
Then , we have
$v=(-0.6)\sec^2 (0)=-0.6$ km/sec
Thus, the speed is $0.6$ km/sec
Since, 1 rev=2 $\pi$ radians
Thus, $0.6 rad/sec=(\dfrac{0.6 rad}{s}) (\dfrac{1 rev}{2 \pi rad}) (\dfrac{60 sec}{1 min}) =\dfrac{18}{\pi}rev/min $
Hence, $0.6$ km/sec and $\dfrac{18}{\pi}rev/min $
