• Support PF! Buy your school textbooks, materials and every day products Here!

A Hard kinematic problem

  • Thread starter Buddy J.
  • Start date
  • #1
5
0

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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
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.
 

Related Threads for: A Hard kinematic problem

Replies
9
Views
979
Replies
6
Views
2K
Replies
60
Views
1K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
8
Views
8K
Replies
31
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
4
Views
3K
Top