# Max distance betw 2 retarding bodies moving towards each other,so that they meet.

1. Apr 6, 2008

### Mr Virtual

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

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

3. The attempt at a solution

Let s1 = Distance travelled by object1 before it stops at last
and s2=Distance travelled 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!

2. Apr 7, 2008

### Dr. Jekyll

$$t=\frac{v_{1}}{a_{1}}=\frac{v_{2}}{a_{2}},$$
$$s_{1}=v_{1}t-\frac{a_{1}}{2}t^{2}=\frac{v_{1}^{2}}{a_{1}}-\frac{v_{1}^2}{2a_{1}}=\frac{v_{1}^{2}}{2a_{1}},$$
$$s_{2}=v_{2}t-\frac{a_{2}}{2}t^{2}=\frac{v_{2}^{2}}{a_{2}}-\frac{v_{2}^2}{2a_{2}}=\frac{v_{2}^{2}}{2a_{2}},$$
$$\frac{(v_{1}+v_{2})^{2}}{2(a_{2}+a_{2})}.$$