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.​

Step-by-step explanation:

(a)

The acceleration is,

`\begin{gathered} a=(dv)/(dt) \\ =(dv)/(dx)*(dx)/(dt) \\ =v(dv)/(dx) \end{gathered}`

According to the question,

`\begin{gathered} v(dv)/(dx)=-4x \\ vdv=-4xdx \\ \int vdv=\int-4xdx \\ (v^2)/(2)=-(4x^2)/(2)+c \end{gathered}`

Here, 'c' is the integration constant.

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

`\begin{gathered} ((4)^2)/(2)=-(4*0)/(2)+c \\ (16)/(2)=c \\ c=8 \end{gathered}`

Now, the velocity is,

`\begin{gathered} (v^2)/(2)=-(4x^2)/(2)+8 \\ v^2=-4x^2+16 \\ v=√(16-4x^2) \end{gathered}`

Hence, the velocity as a function of x is,

`v=√(16-4x^2)`

(b)

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

So, the maximum position is,

`\begin{gathered} 0=√(16-4x^2) \\ 16-4x^2=0 \\ 4x^2=16 \\ x^2=(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=(dx)/(dt)=√(16-4x^2) \\ dx=√(16-4x^2)dt \\ (dx)/(√(16-4x^2))=dt \\ \int(dx)/(√(16-4x^2))=\int dt \\ (1)/(2)sin^(-1)((x)/(2))+k=t \end{gathered}`

Here, 'k' is the integration constant.

Hence, the position as a function of time is,

`(1)/(2)sin^(-1)((x)/(2))+k=t`