How Do You Solve a Kinematic Problem with Time-Varying Acceleration?

[Buddy J.]

Homework Statement

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'')

Homework Equations

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

(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?

 

[Dick]

Your basic given equation is a(t)=dv(t)/dt=A/t^2 for some A. So integrate that to get expressions for v(t) and x(t) being sure to keep constants of integration. You'll get two of them. So now you have three unknown constants and three 'boundary conditions'. So you should be able to solve for all of the constants and answer any questions you need to.