I think it’s the first one
Answer:v=41.23 m/s
Explanation:
Given
mass of heavy object 
distance of
from the axle 
mass of rock 
Length of rod 
distance of
from axle 
Net torque acting is




Work done by
is converted to rock kinetic Energy
thus

Where 





V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
Answer:
jwuwhisbuebeu said that good morning and private s the
This seems like an incomplete question..