Two particles P and Q simultaneously start moving from point A with velocities $15\,m{{s}^{-1}}$ and $20\,m{{s}^{-1}}$ respectively. The two particles move with accelerations equal in magnitude but opposite in direction. When P overtakes Q at B then Its velocity is $30\,m{{s}^{-1}}$. The velocity of Q at point B will
A. $30\,m{{s}^{-1}}$
B. $5\,m{{s}^{-1}}$
C. $10\,m{{s}^{-1}}$
D. $15\,m{{s}^{-1}}$

Hint:Here Newton’s equation of motion is used. It is given that two particles start from the same position. They have the same acceleration but in opposite directions. The velocity of P when it overtakes Q is given. From this we can apply Newton's equation of motion to find out the velocity of Q at the point B.

Formula used:
Newton’s equations of motion are used.
1.$v=u+at$
2.$s=ut+\frac{1}{2}a{{t}^{2}}$
3.${{v}^{2}}={{u}^{2}}+2as$
where v= final velocity
u=initial velocity
a= acceleration
t= time
s= distance travelled

Complete step by step solution:
It is given that both particles simultaneously start from the same point but with different velocities.
Initial velocity of particle P, $v=15\,m{{s}^{-1}}$
Initial velocity of particle Q, \[v=20\,m{{s}^{-1}}\]

Let ‘a’ be the acceleration of particle P and ‘-a’ be the acceleration of particle Q. Then by applying Newton’s first equation of motion to particle P at the position B, we get:
Velocity, $30=15+at$
That is, $at=15$

Now consider velocity of particle Q at the position B be v. Now applying Newton’s first equation of motion to particle Q at the position B, we get:
Velocity of Q at B, \[v=20-at\]
On substituting, we get:
Velocity of Q at B,
\[v=20-15=5\,m{{s}^{-1}}\]

Hence, the answer is option B.

Note: We can easily solve it if we know how to apply Newton’s equation of motion. Don’t get confused with the concept of relative velocity. Negative acceleration is taken for Q because acceleration is a vector quantity and has both direction and magnitude and it is given that both particles have opposite acceleration.