1. The problem statement, all variables and given/known data if you knew that the acceleration of a body is inversely proportional with square the time from the time interval (t=2'' to t=10''), and at t=2'' its velocity was V=-15 m/sec., and at t=10'' its velocity was V=-0.36 m/sec. if its postion from the origin point at t=2'' is twice as much at t=10'', find 1-the position of the body at t=2'' and at t=10'' 2-the distance covered from the time interval (t=2'' to t=10'') 2. Relevant equations a=v(dv/dx) or dv/dt........v=dx/dt........D=r-r'..........r is the position vector 3. The attempt at a solution i put the relation in the form (a2/a10)=(t²10/t²2) i got the relation that a2=25a10 then i integrated to get that integration (dv2/dt)=25 integration (dv10/dt) V2=25V10+c then i used the velocities were given to me to get c=-6...... then i couldn't go on, because i don't know what i am getting and what for can you help me please?