1
$\begingroup$

Suppose the acceleration of a particle is a function of $x$, where

$$a(x) = (2.2 s^{-2})x$$

(a) If the velocity is zero when x = 1.0 m, what is the speed when x = 3.4 m?

(b) How long does it take the particle to travel from x = 1.0 m to x = 3.4 m?

So far for part a, I have tried integrating a(x) to get v(x) = (1.1s^-1)x^2-1.1. I got the -1.1 constant because the velocity is zero at x=1m. This isn't right however because it is dimensionally incorrect.

For part b, I'm not sure how to convert the equations into time, or if I even need to convert the equations. Please explain how to do parts a and b!

$\endgroup$
  • 2
    $\begingroup$ Welcome New contributor user687319! I recommend that you edit your question for readability by using mathjax for the equations. I've made a partial edit to get you started. $\endgroup$ – Alfred Centauri Sep 25 at 1:06
0
$\begingroup$

Substitute $a$ as $v {dv\over dx}$. You can then take $dx$ to the other side and integrate to get velocity in terms of x. As we know $v=0$ when $x=2$, we can use this to find the constant of integration.The resultant equation will solve (a).

You can again write $v$ as ${dx\over dt}$ and integrate to get x in terms of t. Use this to solve (b).

$\endgroup$
3
$\begingroup$

Consider that the work done is converted into kinetic energy

$$ \frac{1}{2} m v^2 = \int F {\rm d}x $$ where $F = m a$.

$$ \frac{1}{2} v^2 = \int a\,{\rm d}x $$

You go from there...

$\endgroup$
0
$\begingroup$

Hint:) Write a=vdv/dx=2.2$x$, integrate it in above limit, so net displacement is 2.4, until velocity is zero, so put it in S=ut+1/2at^2

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.