Question

The acceleration-displacement graph of a partice moving in a straight line is shown in the figure. Initial velocity of the particle is zero. Find the velocity of the particle when displacement of the particle is $12\text{m}$.

• A. $3\sqrt{2}\text{m/s}$
• B. $2\sqrt{5}\text{m/s}$
• C. $4\text{m/s}$
• D. $4\sqrt{3}\text{m/s}$

Answer 1

The correct option is D $4\sqrt{3}\text{m/s}$

The area under $a-S$ graph gives $\frac{v^2-u^2}{2}$,

Thus, $\int adS=\frac{v^2-u^2}{2}...\left(1\right)$

Given,
Initial velocity, $u=0$


From the given figure,

$\int adS=\text{Area of}\Delta \text{OAE+Area of rectangle ABFE+ Area of trapezium FGCB + Area of}\Delta \text{CGD}$

Now, substituting the values from graph

$\int adS=(\frac{1}{2}\times 2\times 2)+[2\times (8-2\left)\right]+\frac{1}{2}(10-8)(2+4)+\frac{1}{2}(12-10)4$

$\Rightarrow \int adS=2+12+6+4=24...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get

$\frac{v^2-0^2}{2}=24$

$\Rightarrow \frac{v^2}{2}=24$

$\Rightarrow v=\sqrt{48}$

$\therefore v=4\sqrt{3}\text{m/s}$

Hence, option (d) is the correct answer.