(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# A Hard kinematic problem

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