Question

Two particles having position vectors r1=(3i+5j)m r2= (-5j+3j)m are moving with velocities v1 = (4i-4j )m/s and v2=(-ai+3j)m/s .if they collide after 2 seconds ,the value of 'a' is

Answer 1

In this question, Second Equation of Motion is to be used and also in this case the acceleration is zero, a=0
two objects collide when their position vector is same.

Also, I have used s instead of r.

Now, You see, This Question is Wrong, Two objects moving under given conditions will never collide.

But Let's just say that you have misread this question.
Now The Question can be rectified only if value of v₂ is

             v₂= (-ai - 3j)

Then The answer would be a= -8

You can figure that out for Yourselves

Thank U, for Your Time.

Answer 2

Please try to post the question with correct values. Because as per the given data the 2 particles will never collide .

Let us consider the values as following:
r1=(3i +5j) m
r2=(-5i+3j) m
v1=(4i - 4j) m/s
 and let v2 be = (-ai-3j)m/s
t= 2 sec

a=:?

If two particles collide after time "t" they travel equal distance .

r1+v1t=r2+v2t

(3i+5j)+(4i-4j)x2= -5i+3j+(-ai-3j)2

on solving we get,

3i+8i+5j -8j= -5i +3j -2ai -6j

11i +5i -3j = -2ai -3j

16i=- 2ai

a= -16/2
=- 8m/s²

∴a= -8m/s² 

Hope the solution has helped you.Thanks.