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'')
a=v(dv/dx) or dv/dt...v=dx/dt...D=r-r'...r is the position vector
The Attempt at a Solution
i put the relation in the form
i got the relation that a2=25a10
then i integrated to get that
integration (dv2/dt)=25 integration (dv10/dt)
then i used the velocities were given to me to get
then i couldn't go on, because i don't know what i am getting and what for
can you help me please?