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.

 

[m2003]

As a scientist, it is important to approach problems with a clear understanding of the variables involved and the equations that can be used to solve for them. In this case, the given information includes the acceleration of the body, the initial and final velocities at two different time intervals, and the position vector. It is also important to note that the acceleration is inversely proportional to the square of the time interval, which can be expressed as a2=25a10.

To solve for the position of the body at t=2'' and t=10'', we can use the equations of motion, specifically the equation for displacement (D=r-r'), where r is the final position and r' is the initial position. We can also use the fact that velocity is the derivative of position (v=dx/dt).

At t=2'', the initial position is not given, so we can use the equation v=dx/dt to find the position at this time. Plugging in the given velocity of V=-15 m/sec and the time t=2'', we get:

-15=dx/dt *2

dx/dt=-7.5 m/sec

Integrating this, we get the position at t=2'' to be:

x=-7.5t + c

At t=10'', the final position is given to be twice the initial position, so we can write:

r=2r'

Using the equation for velocity, we can also write:

v=dx/dt=2(dx/dt)

Substituting this into the equation v=a(t), we get:

2(dx/dt)=-a(t)

Integrating both sides, we get:

2x=-a(t)^2 + c

At t=10'', we can plug in the given velocity of V=-0.36 m/sec and the time t=10'':

-0.36=2x/10^2 + c

Solving for c, we get:

c=-0.36 + 2x/10^2

Substituting this into the equation 2x=-a(t)^2 + c, we get:

2x=-a(t)^2 -0.36 + 2x/10^2

Solving for x, we get:

x=-a(t)^2/2 + 0.36/2 + x/10^2

x=-a(t)^2/2 + 0