Question

Acceleration of a particle moving on a straight line is a=-4x. Where x is position when the particle was at origin it has velocity of +4(m/s)a) Find velocity as function of xb) Maximum position of the particlec) Position as function of time.

Answer 1

Given:

The acceleration of the particle is,

$a=-4x$

The particle was in origin with velocity,

$v_o=+4\text{ m/s}$

To find:

a) Find velocity as a function of x

b) Maximum position of the particle

c) Position as a function of time.​

Explanation:

(a)

The acceleration is,

$\begin{gathered} a=\frac{dv}{dt} \\ =\frac{dv}{dx}\times\frac{dx}{dt} \\ =v\frac{dv}{dx} \end{gathered}$

According to the question,

$\begin{gathered} v\frac{dv}{dx}=-4x \\ vdv=-4xdx \\ \int vdv=\int-4xdx \\ \frac{v^2}{2}=-\frac{4x^2}{2}+c \end{gathered}$

Here, 'c' is the integration constant.

Now, applying the condition at the origin, we can write,

$\begin{gathered} \frac{(4)^2}{2}=-\frac{4\times0}{2}+c \\ \frac{16}{2}=c \\ c=8 \end{gathered}$

Now, the velocity is,

$\begin{gathered} \frac{v^2}{2}=-\frac{4x^2}{2}+8 \\ v^2=-4x^2+16 \\ v=\sqrt{16-4x^2} \end{gathered}$

Hence, the velocity as a function of x is,

$v=\sqrt{16-4x^2}$

(b)

At the maximum position, the speed of the particle becomes zero.

So, the maximum position is,

$\begin{gathered} 0=\sqrt{16-4x^2} \\ 16-4x^2=0 \\ 4x^2=16 \\ x^2=\frac{16}{4} \\ x^2=4 \\ x=\pm2 \end{gathered}$

Hence, the position of the particle is,

$x=\pm2\text{ m}$

(c)

Now, the velocity of the particle is,

$\begin{gathered} v=\frac{dx}{dt}=\sqrt{16-4x^2} \\ dx=\sqrt{16-4x^2}dt \\ \frac{dx}{\sqrt{16-4x^2}}=dt \\ \int\frac{dx}{\sqrt{16-4x^2}}=\int dt \\ \frac{1}{2}sin^{-1}(\frac{x}{2})+k=t \end{gathered}$

Here, 'k' is the integration constant.

Hence, the position as a function of time is,

$\frac{1}{2}sin^{-1}(\frac{x}{2})+k=t$