Question

A particle is kept at rest at origin. Another particle starts foleom 5m
,0 with a velocity of - 4i+3j m/s find their closet distance of approach

Answer 1

Answer: 3 m

Explanation:

1)

Let the the position of first particle be O(0,0) and other one be A(5,0) .

The 2nd particle at A moves with velocity -4i + 3j whose direction is shown in figure.

Since,  this velocity make line whose slope is negative ,also passes through  A.

2) By velocity,  

Slope of line :

m = -3/4 .

Equation of line,  

$(y -0) = \frac{-3}{4} (x-5) \\ \\=> 3x+4y=15$

3) Closest point of approach from origin is the perpendicular distance from O to that line,  which meets at some point C.

In ΔAOB,  

OB = 15/4m

OA = 5 m

By using Pythagoras Theorem,  

$AB^2=OB^2+OA^2 \\ \\=>AB^2 = (15/4)^2+5^2\\ \\=>AB = 25/4$

4) We can write area of ΔAOB in two ways :

$\frac{1}{2}*OA*OB=\frac{1}{2}*OC*AB\\ \\=> 5*15/4=OC*25/4\\ \\=>OC=3 m$

Hence,  closest point of approach is 3m .