Question

From the $\text{velocity-time}$ graph, as given in the figure, of a particle moving in a straight line, one can conclude that,

• A. its velocity for the first $3\text{s}$ is uniform and is equal to $4{\text{ms}}^{-1}$.
• B. The body has a uniform retardation from $t=8\text{s}$ to $t=12\text{s}$
• C. its average velocity during the $12\text{s}$ interval is $\frac{24}{7}{\text{ms}}^{-1}$.
• D. the body has a non-zero constant acceleration between $t=3\text{s}$ and $t=8\text{s}$.

Answer 1

The correct option is B The body has a uniform retardation from $t=8\text{s}$ to $t=12\text{s}$

$\text{Displacement in}12\text{s}=\text{area under}v-t\text{graph}$

$=\frac{1}{2}\times (12+5)\times 4=34\text{m}$

We know that,

$V_{av}=\frac{\text{Displacement}}{\text{Time}}=\frac{34}{12}=\frac{17}{6}{\text{ms}}^{-1}$

Hence, $\left(\text{a}\right)$ is incorrect.

Option $\left(\text{b}\right)$ is incorrect, because during first $3\text{s}$, velocity increases from $0$ to $4\text{m/s}$.

Option $\left(\text{c}\right)$ is incorrect, because in part $\text{AB}$, velocity is constant due to which acceleration becomes $0$.

From $t=8\text{s}$ to $t=12\text{s}$, slope is negative due to which the body is having uniform retardation.

Hence, option $\left(\text{D}\right)$ is the correct answer.