Question

The acceleration of a particle is defined by the relation, $a=-4x(-1+\frac{1}{4}{x}^2)$ , where $x$ is displacement along $x-$ axis. All the quantities are in $SI$ units. If velocity of the particle $\left(v\right)=17\text{m/s}$ when $x=0$, then the velocity of the particle when $x=4\text{m}$ is

• A. $12\text{m/s}$
• B. $15\text{m/s}$
• C. $20\text{m/s}$
• D. $25\text{m/s}$

Answer 1

The correct option is B $15\text{m/s}$

Given,
$a=-4x(-1+\frac{1}{4}{x}^2)=v\frac{dv}{dx}$
$\Rightarrow {\int }_{17}^{v}vdv={\int }_0^4(4x-x^3)dx$
$\Rightarrow \frac{1}{2}[v^2{]}_{17}^v=\frac{4}{2}[x^2{]}_0^4-\frac{1}{4}[x^4{]}_0^4$
$\Rightarrow v^2-{17}^2=4(4^2-0)-\frac{1}{2}(4^4-0)$
$\Rightarrow v^2=225$
$\Rightarrow v=15\text{m/s}$