Question

The figure shows velocity-time graph of a particle moving along a straight line. Identify the correct statement:

• A. The particle crosses its initial position at $t=3\text{s}$.
• B. The particle crosses its initial position at $t=2\text{s}$.
• C. The average speed of the particle in the time interval, $0\le t\le 2\text{s}$ is zero.
• D. Initial speed of particle is zero.

Answer 1

The correct option is B The particle crosses its initial position at $t=2\text{s}$.

$\text{Area under}v-t\text{graph}=\text{Displacement}$
Displacement in time interval $0\text{to}1\text{s}$
$=\text{Area}\left(\Delta OAB\right)$
$=\frac{1}{2}\times 1\times 10=5\text{m}...\left(1\right)$


Displacement in time interval $1\text{to}2\text{s}$
$=\text{Area}\left(\Delta BCD\right)$
$=-\frac{1}{2}\times 1\times 10=-5\text{m}...\left(2\right)$
Now, net displacement (sum of total area under curve) from $t=0$ to $2\text{s}$ is,
$S=+5+(-5)=0$
$\therefore$ Particle will cross its initial position at $t=2\text{s}$, since net displacement of particle is zero at the end of $t=2\text{s}$.
Total distance covered in $0\le t\le 2\text{s}$ is not zero. So, average speed is also not zero.
And initial speed of particle is not zero. (from the graph)
Hence, only option (b) is correct.