Question

The velocity of a particle moving on the $x$-axis is given by $v=x^2+x$, where $v$ is in $m/s$ and $x$ is in $m$. Find its acceleration (in $m/s^2$) when passing through the point $x=2m$.

Answer 1

Given:
$v=x^2+x$
Acceleration of the particle is given by,
$a=\frac{dv}{dt}...\left(i\right)$
As we know,
$v=\frac{dx}{dt}$
Putting the value of 𝑣 in equation $\left(i\right)$ we get,
$a=v\frac{dv}{dx}\dots .\left(ii\right)$
Therefore, by putting the values in $\left(ii\right)$ equation we get,
$a=(x^2+x)\frac{d(x^2+x)}{dx}$
$=(x^2+x)(2x+1)$
$a=2x^3+2x^2+x^2+x$
$a=2x^3+3x^2+x$
Acceleration of the particle when $x=2m$ is
$a=2(2{)}^3+3(2{)}^2+2$
$=16+12+2$
$a=30m/s^2$