Two particles Aand B move with velocities v, andv, respectively along

1.	Two particles Aand B move with velocities v, andv, respectively along the x & y axis. The initialseparation between them is 'd' as shown in thefig. Find the least distance between them duringtheir motion.d.V2(2) v +V3d.v2d.v2
Answer» 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+v1tThat of b : y = q + v2 tDistance ab :ab^2 = x^2+y^2= p^2+q^2+(v1^2+v2^2)t^2+2t (pv1+qv2)Differentiate wrt t. And set to 0.=> 2t(v1^2+v2^2) + 2(pv1+qv2) = 0t = - (pv1 + q* v2) / (v1^2+v2^2)As t>0, v1 or v2 or both must be negative. => Minimal ab^2 = d^2 - (pv1+qv2)^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. SoMinimal ab = d*v1/sqrt(v1^2+v2^2)Similarly, if q=0, p= d.

Two particles Aand B move with velocities v, andv, respectively along the x & y axis. The initialseparation between them is 'd' as shown in thefig. Find the least distance between them duringtheir motion.d.V2(2) v +V3d.v2d.v2

Answer»

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*tThat of b : y = q + v2 * tDistance 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) = 0t = - (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. SoMinimal ab = d*v1/sqrt(v1^2+v2^2)Similarly, if q=0, p= d.

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