Question

The acceleration of a particle which moves along the positive x-axis varies with its position as shown. The velocity of the particle is $0.8m/s$ at $x=0$ and the acceleration becomes zero after $x=1.4m$. The maximum velocity of the particle (in m/s) is

• A. $1.6m/s$
• B. $1.2m/s$
• C. $1.4m/s$
• D. $1m/s$

Answer 1

The correct option is B $1.2m/s$

As we know that the area under $a-x$ curve

$\left(a=v\frac{dv}{dt}\Rightarrow {\int }_u^{v}vdv\right)$

gives $\left(\frac{v^2-u^2}{2}\right)$ value
So, area of the given curve is
$=0.4\times 0.2+0.4\times 0.2+0.4\times 0.2+\frac{1}{2}\times 0.4\times 0.2+0.6\times 0.2$
$\Rightarrow$ Area $=0.4$
Now, $v^2-u^2=2$ (Area)
And, as per given question $u=0.8m/s$
$\Rightarrow v^2=2\times 0.4+(0.8{)}^2=1.44$
$\Rightarrow v=1.2m/s$
Hence, the maximum velocity of the particle is $1.2m/s$