Question

two particles a and b move with velocities v1 and v2 respectively along the x and y axis. the initial separation between them is d . find the least distance between them during their motion

Answer 1

If particles a and b are moving in positive x and positive directions respectively, then the least distance will be d.

Let a be initially at point (p, 0). Let b be at point (0, q) at t=0. Then d^2 = p^2+q^2.
Position of a at time t : x = p+v1*t
That of b : y = q + v2 * t
Distance ab :
ab^2 = x^2+y^2
= p^2+q^2+(v1^2+v2^2)*t^2+2t *(p*v1+q*v2)
Differentiate wrt t. And set to 0.
=> 2t*(v1^2+v2^2) + 2(p*v1+q*v2) = 0
t = - (p*v1 + q* v2) / (v1^2+v2^2)
As t>0, v1 or v2 or both must be negative.

=> Minimal ab^2 = d^2 - (p*v1+q*v2)^2 / (v1^2+v2^2)
.
The solution depends on initial positions of a and b.

Let a be at origin. Then p=0. q=d. So
Minimal ab = d*v1/sqrt(v1^2+v2^2)
Similarly, if q=0, p= d.

Answer 2

Answer:

Explanation: This is a good question.

It can be solved by the concept of relative motion. Please see the attached file.