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:
$$\frac{(v_{1}+v_{2})^{2}}{2(a_{2}+a_{2})}.$$

 

[Moetasim]

I would approach this problem by first analyzing the given information and the equations involved. From the given information, we know that two bodies are moving towards each other with initial velocities and constant retardation. This indicates that the bodies are decelerating towards each other, and eventually, they will come to a stop when they meet.

In order for the bodies to meet during their motion, the total distance traveled by both bodies must be equal. This can be represented by the equation:

s1 + s2 = d

Where s1 and s2 are the distances traveled by the two bodies, and d is the initial separation between them.

Using the equations of motion, we can rewrite this equation as:

sqr(v1)/2*a1 + sqr(v2)/2*a2 = d

Now, we need to find the maximum value of d for which the two bodies will meet. This can be done by finding the point of intersection of the two curves represented by these equations.

To do this, we can use the quadratic formula to solve for d:

d = (-b ± sqrt(b^2-4ac))/2a

Where a = a1+a2, b = (v1+v2), and c = 0.

Plugging in the values, we get:

d = (-v1-v2 ± sqrt((v1+v2)^2))/2(a1+a2)

Simplifying this further, we get:

d = (-v1-v2 ± (v1+v2))/2(a1+a2)

Since we are looking for the maximum value of d, we take the positive sign and simplify further to get:

d = (v1+v2)/2(a1+a2)

This matches with option b) in the given options. Therefore, the maximum initial separation between the two bodies for which they will meet during their motion is (v1+v2)/2(a1+a2).

In conclusion, as a scientist, I would use the equations of motion and the concept of equal distances traveled to analyze and solve this problem. I would also double check my calculations to ensure that my answer matches with the given options or any other known solution.