Question

The acceleration of a particle which moves along the positive $\text{x-}$axis varies with its position as shown in the figure. If the velocity of the particle is $0.8\text{m/s}$ at $x=0$, the velocity of the particle at $x=1.4$ is (in $\text{m/s}$)

• A. $1.6$
• B. $1.2$
• C. $1.4$
• D. none

Answer 1

The correct option is B $1.2$

We know that, $$a=\frac{vdv}{dx}$$ $\Rightarrow adx=vdv$

On integrating the above equation using given limits, we get

${\int }_{x=0}^{x=1.4}adx={\int }_u^{v}vdv$

${\int }_{x=0}^{x=1.4}adx=\frac{v^2-u^2}{2}...\left(1\right)$

The left hand side $\int adx$ represents area under the curve on ${}^{\prime }{x}^{\prime }$ axis.

Area under curve from $x=0$ to $x=1.8$ will be,

${\int }_{x=0}^{x=1.4}adx=0.4\times 0.4+\frac{1}{2}(0.8-0.4)(0.4-0.2)+(1.4-0.4)(0.2-0)$

$=0.16+\mathrm{0.0.04}+0.2$

$=0.40$

Substituting it in equation $\left(1\right)$, we get

$0.40=\frac{v^2-u^2}{2}$

Given, $u=0.8\text{m/s}$

$\Rightarrow 0.80=v^2-(0.8{)}^2$

$\Rightarrow v^2=0.80+0.64$

$\Rightarrow v^2=1.44$

$\Rightarrow v=\sqrt{1.44}$

$\therefore v=1.2\text{m/sec}$

Hence, option (b) is the correct answer.