Question

The acceleration of a particle is give by a=x, where x is position. If the particle starts at origin from rest, find its distance from origin after time t

Answer 1

Answer:

The distance from the origin will be x.

Explanation:

$x = \frac{1}{2} a {t}^{2}$

Answer 2

Here the acceleration of the particle is numerically equal to its position with time.

$\longrightarrow\sf{a=x}$

Since acceleration is the first derivative of displacement,

$\longrightarrow\sf{\dfrac{dv}{dt}=x}$

Or,

$\longrightarrow\sf{\dfrac{dv}{dx}\cdot\dfrac{dx}{dt}=x}$

Here the term $\sf{\dfrac{dx}{dt}}$ is the rate of change of position of the particle, i.e., its velocity. Then,

$\longrightarrow\sf{v\cdot\dfrac{dv}{dx}=x}$

$\longrightarrow\sf{v\ dv=x\ dx}$

Integrating,

$\displaystyle\longrightarrow\sf{\int v\ dv=\int x\ dx}$

$\displaystyle\longrightarrow\sf{\dfrac{v^2}{2}=\dfrac{x^2}{2}}$

$\displaystyle\longrightarrow\sf{v^2=x^2}$

$\displaystyle\longrightarrow\sf{v=\pm x}$

Since velocity is the first derivative of displacement.

$\displaystyle\longrightarrow\sf{\dfrac{dx}{dt}=\pm x}$

$\displaystyle\longrightarrow\sf{\pm\,\dfrac{1}{x}\ dx=dt}$

Integrating,

$\displaystyle\longrightarrow\sf{\pm\int\dfrac{1}{x}\ dx=\int dt}$

$\displaystyle\longrightarrow\sf{\pm\ln|x|=t}$

$\displaystyle\longrightarrow\sf{\ln|x|=\pm t}$

Here $\sf{t\geq0}$ and $\sf{\ln|x|\ \textgreater\ 0.}$ Then,

$\displaystyle\longrightarrow\sf{\ln|x|=t}$

$\displaystyle\longrightarrow\sf{\underline{\underline{x=e^t}}}$