Question

The figure shows a velocity-time graph of a particle moving along a straight line.


Find the maximum displacement of the particle.

• A. $33.3\text{m}$
• B. $23.3\text{m}$
• C. $18.3\text{m}$
• D. zero

Answer 1

The correct option is A $33.3\text{m}$

As, we know that,
$$v=\frac{dx}{dt}$$ $\Rightarrow \int dx=\int vdt$

Here, $\int dx$ represents net displacement and $\int vdt$ is the area under the $\text{v-t}$curve with sign.

So, for maximum displacement, we should take the points that are on positive side. Hence, area of the shaded region $\text{OABXO}$.

$\text{Area of OABXO}=\frac{1}{2}\times (2-0)\times 10+(4-2)\times 10+\frac{1}{2}\times (X-4)\times 10$

$\Rightarrow \text{Area of OABXO}=10+20+\frac{1}{2}\times (X-4)\times 10...\left(1\right)$


To calculate $X$, using the similarity principle between the $\Delta BGX$ & $\Delta DCX$.

So, $\frac{GX}{GB}=\frac{CX}{DC}$

$\Rightarrow \frac{X-4}{10}=\frac{6-X}{20}$

$\Rightarrow 2X-8=6-X$

$\Rightarrow X=\frac{14}{3}$

Substituting the value of $x$ in equation $\left(1\right)$,

$\Rightarrow \text{Area of OABXO}=10+20+\frac{1}{2}\times (\frac{14}{3}-4)\times 10$

$\Rightarrow \text{Area of OABXO}=33.3$

Therefore, the maximum displacement is$=33.3\text{m}$

Hence, option (a) is the correct answer.