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Acceleration in 2D

Two particles...

Question

Two particles having position vectors $\to r = (3 \to i + 5 j) m$ and ${\to r}_{2} = (- 5 i + 3 j) m$ are moving with velocities ${\to V}_{1} = (4^i - 4^j) m s^{- 1}$ and ${\to V}_{2} = (a^i - 3^j) m s^{- 1}$ . If they collide after $2$ seconds , the value of $a$ is

A

-

2

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B

-

4

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C

-

6

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D

-

8

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Solution

The correct option is D - $8$
Let us consider the values as following:
$\begin{matrix} r_{1} = (3 i + 5 j) m r_{2} = (- 5 i + 3 j) m v_{1} = (4 i - 4 j) m / s a n d l e t v_{2} b e = (- a i - 3 j) m / s t = 2 sec \end{matrix}$
If two particles collides after time $" t "$ they travel equal distance.
$\begin{matrix} r_{1} + v_{1} t = r_{2} + v_{2} t (3 i + 5 j) + (4 i - 4 j) \times 2 = - 5 i + 3 j + (- a i - 3 j) 2 o n s o l v i n g w e g e t, 3 i + 8 i + 5 j - 8 j = - 5 i + 3 j - 2 a i - 6 i 11 i + 5 i - 3 j = - 2 a i - 3 j 16 i = - 2 a i a = \frac{- 16}{2} = - 8 m / s^{2} \end{matrix}$

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