Max Initial Separation for Meeting of Two Moving Bodies

[Mr Virtual]

Homework Statement

Two bodies move in a straight line towards each other at initial velocities v1 and v2 and with constant retardation a1 and a2 respectively at the initial instant. What is the max initial separation between the bodies for which they will meet during the motion?

(sqr -> square of , root ->square root of)
Options:
a) sqr(v1)/a1 + sqr(v1)/a2
b) sqr(v1+v2)/2(a1+a2)
c)v1*v2/root(a1*a2)
d)sqr(v1)-sqr(v2)/(a1-a2)

Homework Equations

sqr(v) = sqr(u) + 2as

The Attempt at a Solution

Let s1 = Distance traveled by object1 before it stops at last
and s2=Distance traveled by object 2 before it stops

0 = sqr(v1) - 2*a1*s1
s1= sqr(v1)/2*a1
s2=sqr(v2)/2*a2

Max dist=s1 + s2

However, this answer does not match with any of the options above. According to the book, the correct answer is option b.
Help!

 

[Dr. Jekyll]

My answer doesn't match too.

I did it like this:

$$t=\frac{v_{1}}{a_{1}}=\frac{v_{2}}{a_{2}},$$

$$s_{1}=v_{1}t-\frac{a_{1}}{2}t^{2}=\frac{v_{1}^{2}}{a_{1}}-\frac{v_{1}^2}{2a_{1}}=\frac{v_{1}^{2}}{2a_{1}},$$

$$s_{2}=v_{2}t-\frac{a_{2}}{2}t^{2}=\frac{v_{2}^{2}}{a_{2}}-\frac{v_{2}^2}{2a_{2}}=\frac{v_{2}^{2}}{2a_{2}},$$

but the sum of the last two expressions doesn't match with the (b) option, for which you say that equals to:
$$\frac{(v_{1}+v_{2})^{2}}{2(a_{2}+a_{2})}.$$