Question

Two particles $\text{P}$ and $\text{Q}$ initially separated by $400\text{m}$, are moving towards each other along a straight line as shown in figure with $a_p=2{\text{m/s}}^2,u_p=15\text{m/s}$ and $a_Q=4$${\text{m/s}}^2,u_Q=25\text{m/s}$. Calculate the time when they meet.

• A. $\frac{20}{3}\text{s}$
• B. $\frac{40}{3}\text{s}$
• C. $\frac{50}{3}\text{s}$
• D. $\frac{70}{3}\text{s}$

Answer 1

The correct option is A $\frac{20}{3}\text{s}$

Considering ground as frame of reference and choosing the origin at $\text{P}$ and right as positive.


Given,

$u_P=15\text{m/s}$	$u_Q=25\text{m/s}$
$a_P=2{\text{m/s}}^2$	$a_Q=-4{\text{m/s}}^2$

Now, relative velocity and acceleration of point $\text{Q}$ with respect to point $\text{P}$.

$\overrightarrow{u_{QP}}=\overrightarrow{u_Q}-\overrightarrow{u_P}=-25-(+15)=-40\text{m/s}$

$\overrightarrow{a_{QP}}=\overrightarrow{a_Q}-\overrightarrow{a_P}=-4-(+2)=-6{\text{m/s}}^2$

Now, considering $\text{P}$ as observer, choosing the origin $\text{P}$ and right as positive.


So, the velocity and acceleration in this frame are:

$u_{pp}=0\text{m/s}$	$u_{Qp}=-40\text{m/s}$
$a_{pp}=0{\text{m/s}}^2$	$a_{Qp}=-6{\text{m/s}}^2$

Now, applying the equation of motion:

$x=x_0+ut+\frac{1}{2}a{t}^2$

Here,
$s_{QP}=(s_{QP}{)}_0+u_{QP}t+\frac{1}{2}{a}_{QP}{t}^2$

Where, $t$ is the time when they meet.
Substituting the values,

$\Rightarrow 0=400-40t+\frac{1}{2}\times (-6)\times t^2$

$\Rightarrow 3t^2+40t-\left(400\right)=0$

$\Rightarrow 3t^2-20t+60t-\left(400\right)=0$

$\Rightarrow t(3t-20)+20(3t-20)=0$

$\Rightarrow (3t-20t)(t+20)=0$

$t=\frac{20}{3}$ and $t=-20$

Taking positive solution, we have

$t=\frac{20}{3}\text{s}$

Hence, option (a) is the correct answer.