Distance after pushoff between two objects

[spider3367]

Homework Statement

A 60 kg parent and a 12 kg child meet at the center of an ice rink. They push off each other for .5 seconds. If the acceleration of the child is 2.6 m/s^2, then what is the distance between the parent and the child at 3 seconds after their push off?

Homework Equations

F=ma
X=Xo +volt + 1/2at^2

The Attempt at a Solution

I know that the net force of the child is (12)(2.6)= 31.2 N. Using Newtons equal and opposite law, I know that the force of the parent must be -31.2 N. Then to find the acceleration of the parent, -31.2=(60)a which means a= -.52 m/s^2. From there I think you plug in each of of the acceleration with time 2.5 into the equation above to get each of their positions (X=0 + 0 + 1/2 at^2). with that I get about 6 meters difference and I know that is not the answers

 

[tiny-tim]

Hi spider3367!

(try using the X² tag just above the Reply box )

Yes, you've basically done it correctly, but you've not read the question properly!

The acceleration is as given for .5 seconds, but after that, the acceleration is zero.

(also, you could have got .52 m/s² simply by using good ol' Newton's first law … there's no external force, so the position of the center of mass will be stationary)

Try again! ​

 

[spider3367]

So I'm still a little confused.

(2.6 m/s2)(.5^2)*1/2=.325
(.52m/s2)(.5^2)*1/2=.065

add them= .39 meters? That seems like it can't be right.

 

[tiny-tim]

(please use the X² tag just above the Reply box )

i] you could have added the 2.6 to the .52 to get the relative acceleration, abnd put that into the formula

ii] you still have another 2.5 seconds of constant speed to account for.

(and I'm going to bed :zzz:)​

 

[spider3367]

Am I on the right track with this:

xtotal=[0+(2.6)(2.5)] + [0+(.52)(2.5)] + .39 meters (from calculated accelerations)

 

[tiny-tim]

Hi spider3367!

(just got up :zzz: …)

xtotal=[0+(2.6)(2.5)] + [0+(.52)(2.5)]

Nooo … you need the speed after 0.5 seconds.