Question

From the velocity- time graph, given in figure of a particle moving in a straight line, one can conclude that

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

Answer 1

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

Displacement in $12\text{s}$ = area under $v-t$ graph
$=\frac{1}{2}\times (12+5)\times 4=34\text{m}$
$V_{av}=\frac{\text{Displacement}}{\text{Time}}=\frac{34}{12}=\frac{17}{6}{\text{ms}}^{-1}$
Hence (a) is incorrect.
Option (b) is incorrect because during first $3\text{s}$ velocity increases from 0 to $4\text{m/s}$
Option (c) is incorrect, because in part $AB$, 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.