Question

The following plot shows the velocity-time relation of a moving body.
Then, pick the CORRECT statement(s) from the following.

• A. The body decelerated at a rate of $5\text{m}{\text{s}}^{-2}$ in the last two seconds of its motion.
• B. The body is displaced by $62.5\text{cm}$.
• C. The body accelerated at a rate of $5\text{m}{\text{s}}^{-2}$ in the first four seconds of its motion.
• D. The body has uniformly accelerated motion from a total of $0\text{to}4\text{s}$.

Answer 1

Answer: (A), (B)

The body is displaced by $62.5\text{cm}$.

The slope of velocity-time graph of a body gives the acceleration of the body.
Acceleration from time ($\text{t}$) = $3\text{s}$ to $\text{t}=4\text{s}$ is $\frac{10-5}{4-3}=5\text{m}{\text{s}}^{-2}$
$⟹$ The body is accelerating from $\text{t}=3\text{s}$ to $\text{t}=4\text{s}$, i.e. for one second.
And, acceleration from time ($\text{t}$) = $7\text{s}$ to $\text{t}=9\text{s}$ is $\frac{0-10}{9-7}=-5\text{m}{\text{s}}^{-2}$
An increasing slope indicates that the body is accelerating, while a decreasing one indicates deceleration.

Also, from $\text{t}=0\text{s}$ to $\text{t}=3\text{s}$, the slope is zero and from $\text{t}=3\text{s}$ to $\text{t}=4\text{s}$, the slope is non-zero and constant. Hence, the body has non-uniform acceleration for the interval $\text{t}=0\text{s}$ to $\text{t}=4\text{s}$.

The area under the velocity-time graph gives the total (net) displacement of the body:
The sum of the areas of the sections A to E gives the net displacement.

$⟹$ Net displacement = $15\text{cm}+5\text{cm}+2.5\text{cm}$$+30\text{cm}+10\text{cm}=62.5\text{cm}$