The velocity of a particle moving along the x- axis is given for t0 by V =( 32.0 t^2 - 2.0 t^3 ) m/s , where t is in sec .What is the acceleration of the particle when (after t=0)it achieves its maximum displacement in the positive x - direction?

Question

The velocity of a particle moving along the x- axis is given for t>0 by V =( 32.0 t^2 - 2.0 t^3 ) m/s , where t is in sec .What is the acceleration of the particle when (after t=0)it achieves its maximum displacement in the positive x - direction?

badhi · Answer

You can find the displacement and the acceleration by using integration and differentiation of $v(t)$ respectively (take them as $s(t)$ and $a(t)$) . To find the maximum displacement we have to find the maximum value of s(t). Since we equal the differentiation of s(t) to 0 to find the maximum, it will eventually be $v(t)=0$. So we need not to integrate v(t) after all. 

So solve $v(t)=0$, then you'll find the maximum/minimum point of s(t) and the time that happens($t_1,t_2$). Find out which of them is the maximum (take $t_{max}$). Substitute $t_{max}$ with $a(t)$ to find the acceleration at that point.

anonymous · Answer

You can find the equation which will give the displacement of the particle at any instant of time by integrating the equation you gave. For integrating, there are some simple rules I can tell you, these are :
1. If you want to find the integral of sum (or difference) of two functions of 'x' (or any other variable), you can find the sum (or difference ) of the individual integral of function of x.
2. If you want to find the integral of 'x' to the power of 'n' (x^n), its solution will be 'x' to the power of 'n+1' divided by 'n+1'.
GOT IT.
So, in you case, V = 32t^2 - 2t^3
Now when you will integrate both sides with respect to t, you'll get,
Displacement    x = {32t^(3+1)}/3+1     -     {2t^(2+1)}/2+1
                          =  8t^4 - (2t^3)/3 

( Remember that the integral of constant times function of t (or any other variable) is constant times the integral of function of t. It appears that the constant get out of the integral sign.)
Now to find the time when the displacement is maximum, you'll have to do some maths, don't worry if you don't know how to do this, I will tell you if you want to know.
                                              Post 1 Ended.

anonymous · Answer

Now to find the time when displacement is maximum, we have to find the maxima of the displacement time function we obtained by integrating velocity time equation. For this we need to equate it derivative to zero and then solve for 't', what we'll get will be our required value of 't'. Now , as you now that we obtained displacement time equation by integrating velocity time equation, you can easily figure out that the derivative of displacement time equation will be velocity time equation. And when you'll equate it to zero, you'll get,
32t^2 - 2t^3 = 0, and on solving, t = 16.
Now you'll have to put this value of 't' in the derivative of velocity time equation.
On differentiating it, you'll get acceleration a = 64t - 6t^2 (hope you know differentiation)
Substitute t = 16 and you'll get a = -512 m/s^2        (If you are using SI units)

Hope you understood it.

anonymous · Answer

If you have any other query, you can mail me at hitarth.math@gmail.com

anonymous · Answer

Well, not only you, anyone can ask me any thing and I will try to solve their problem, if I know to solve it.