Question

The velocity-time graph of a particle moving along a straight line is shown in the figure given below. The displacement of the particle in 5 seconds is

(A) $0.5m$
(B) $1m$
(C) $2m$
(D) $4m$

Answer 1

Hint: From the graph, we can say that the velocity of the particle is continuously changing. Velocity is the displacement travelled in unit time, from this relation; we know that displacement is the product of time and velocity so it is the area of the graph. Therefore, to calculate the displacement, we calculate the area under the graph.

Formulas used:
$v=\dfrac{d}{t}$
$ar(\Delta OAB)=\dfrac{1}{2}\times b\times h$
$ar(EFGD)=b\times h$

Complete step-by-step solution:
The graph shows relation between velocity and time of a particle. The time is the independent variable and velocity is the dependent variable.

From the graph, we can see that the velocity is constantly changing. Since the graph is a straight line therefore, the velocity is directly proportional to time.
We know that,
 $v=\dfrac{d}{t}$
Here, $v$ is the velocity
$d$ is the displacement
$t$ is the time taken
From the above equation, we have,
$d=v\times t$
Therefore, the total displacement covered by the particle in 5 seconds is the total area of the graph,

From the figure, area of OAB –
$b=3units$, $h=2units$
$\begin{align}
  & ar(\Delta OAB)=\dfrac{1}{2}\times b\times h \\
 & \Rightarrow ar(\Delta OAB)=\dfrac{1}{2}\times 3\times 2 \\
 & \Rightarrow ar(\Delta OAB)=3sq\,units \\
\end{align}$
After point B, the particle resumes motion in the opposite direction. So the displacement cancels out
$ar(AMB)$ and $ar(BCD)$ cancels out.
Hence,
$\begin{align}
  & ar(OAM)=\dfrac{1}{2}\times 2\times 2 \\
 & \Rightarrow ar(OAM)=2sq\,units \\
\end{align}$
Area of EFGD
$b=1unit$, $h=2units$
$\begin{align}
  & ar(EFGD)=b\times h \\
 & \Rightarrow ar(EFGD)=2\times 1 \\
 & \therefore ar(EFGD)=2units \\
\end{align}$
Therefore, the total effective area is
$ar(OAM)+ar(EFGD)=2+2=4sq\,units$
The total effective area of the graph is $4sq\,units$. Therefore, the total displacement covered is $4m$.

Hence, the correct option is (D).

Note:
The quantity taken along x-axis is the independent variable and the quantity taken along y-axis is the dependent variable. The velocity and time graph is a straight line; this means that velocity is directly related to time. As the velocity is continuously changing, the particle undergoes acceleration.