Question

The graph in figure given below represents the velocity-time graph of a particle, starting from rest at point $\text{P}$. Find the time when particle will reach point $\text{P}$ again.

• A. $8\text{s}$
• B. $10\text{s}$
• C. $12\text{s}$
• D. $16\text{s}$

Answer 1

The correct option is C $12\text{s}$


Let after $t$ seconds, particle reaches $\text{P}$ again.
$\Rightarrow \text{Total displacement}=0$
which means the particle has reversed its direction of motion as shown in below figure.


From $v-t$ curve, we find that particle reverses velocity at $t=8s$.
Also acceleration of the particle $|a|=1\text{m/s^2}$ (slope of the graph is 1)

If particle reaches back to point $\text{P}$ at $t$ seconds,
then area of the graph starting from $t=8\text{s}$ must equal the area of the graph from $t=0$ to $t=8\text{s}$

Let $v$ be the velocity of particle at $t$ seconds.

Then $\text{net area of}v-t\text{curve}=0$
$\Rightarrow \frac{1}{2}\times 8\times 2-\frac{1}{2}\times (t-8)\times v=0$....(1)

Using $v=u+at=1\times (t-8)=t-8$ in above Eq.(1):-
we get $(t-8{)}^2=16$
i.e $t-8=4$
Hence, $t=12\text{s}$