A cord passes over a weightless and frictionless pulley. Masses of 200 g and 300 g are attached to the ends of the cord. Find the distance the masses will move during 5th second after they are started from rest?

This is coming from  a16 year old but I think I'm right  "This problem is a Newton's 2nd problem: SF=m*a where SF is the sum of
the external forces acting on the system, m is the total mass
accelerating and a is the rate of acceleration.

In this case the sum of the external forces is equal to the difference
in the weights of the two masses: SF=m1*g-m2*g=(m1-m2)*g while the total
mass accelerating is m1+m2. Therefore, the acceleration of this system
is: A=(m1+m2)/[(m1-m2)*g]" therefore  a=(200+300)/[(200/300)*9.8]=

76.5306122, so the distance traveled in 5 seconds is

76.5306122 + 153.061224 + 229.591837+306.122449+382.653061 which equals 1 147.95918 meters so in total that's how far the masses will move

source: en.allexperts.com