Variable acceleration question

By Anonymous on Thursday, April 5, 2001 - 10:54 am :

Hi,

I can't seem to see how to go about this Mechanics question.

A particle of mass P 1.5kg moves in a straight line through a fixed point O. At time t sec after passing through O the distance of P from O is x cm and the force acting on P has magnitude (3x + 6)N directed away from O. Given that P passes through O with 2.2^0.5 m s^-1 .

Calculate the value of t when x=20.

Using F = ma, and letting a = v(dv/dx), and then integrating and putting in x=0 and the v=2.2^0.5 . I got the equation to be (3/2)x² + 6x + 6 = (3/4)v² . But this only allows you to find v in terms of t. How do you go about getting t in terms of x?

Thanks for you help in advance NRich.

By Kerwin Hui (Kwkh2) on Thursday, April 5, 2001 - 03:46 pm :

One way of doing this is rewrite the equation

(3/2)x² +6x+6=(3/4)v²

as

2(x² +4x+4)=v²

so, since at t=x=0, v is positive, we have

2^½ (x+2)=v=dx/dt

and now you can integrate this to give the answer.

Kerwin

By Anonymous on Thursday, April 5, 2001 - 04:12 pm :

Hi Kerwin,

Thank you for your help.
It worked out fine. Can you confirm that the answer was 1.6955 which to 2dp comes to 1.7 seconds?

By Anonymous on Thursday, April 5, 2001 - 04:13 pm :

If you could not easily square v, then what would be another approach to solving the problem? Or would you take the same route? Is there another way to answer the question?

By Kerwin Hui (Kwkh2) on Thursday, April 5, 2001 - 08:22 pm :

If the force is given in the form

F = F (x)

, then we can always integrate to give

$v^{2} = 2 \int^{x} F (y) dy$

which means

$dx / dt = \sqrt{2 \int^{x} F (y) dy} =$ a function $g (x)$

and can be integrated from there, though numerical methods may be needed.

Kerwin

By Anonymous on Thursday, April 5, 2001 - 10:45 pm :

Thanks Kerwin