Help Jane & John Pull a Rope: Solve Acceleration & Meetup

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John's acceleration is calculated to be 0.756 m/s² using the formula f=ma. Jane and John start 13 meters apart, with Jane accelerating towards John at 0.92 m/s². To determine where they meet, both must travel for the same time interval, which can be calculated using the equation x = 1/2*a*t². The discussion emphasizes the need to find the correct distance each travels before meeting. The calculations suggest Jane will end up closer to John than initially estimated.
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Here is a question:

Jane and John with masses of 51kg and 62kg respectively, stand on a frictionless surface 13m apart. John pulls on a rope that connects him to JAne, giving jane an acceleration of .92 m/s*s toward him. a. What is Johns acceleration?? b. IF the pulling force is applied constantly, where will Jane and John meet?

a. f=ma, manipulating this you get John's Acceleration at .756m/s*s

b. More towards John. I got 5.9m away from him but I believe this is wrong..Please help ASAP
 
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Both are starting from rest simultaneously.When they meet they must have traveled for the same time intrerval. Use the equation x = 1/2*a*t^2 for both and find x
 

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