Question

A particle is kept at rest at origin another particle starts at (4,0) with veloctiy -3i cap, 4j cap find closest distance of approach

Answer 1

Answer:

3.2

Step-by-step explanation:

A particle is kept at rest at origin another particle starts at (4,0) with veloctiy -3i cap, 4j cap find closest distance of approach

Particle starts at (4,0) with veloctiy -3i , 4j  

Position of particle at time t  

= (4 – 3t ) , 4t

Distance from origin² = (4 – 3t – 0)² + (4t – 0)²

Distance from origin² = 16 + 9t² – 24t + 16 t²

Distance from origin² = 25t²  – 24t + 16

On differentiating with t and equating with 0

 50 t – 24 = 0

t= 24/50

t = 12/25

Distance from origin² =  ( 4 – 3*12/25) 2 + (4*12/25) 2

=  ( 64/25) 2 + (48/25) 2

= (1/25)² ( 4096 + 2304)

= (1/25)² ( 6400)

= ( 80/25)²

Distance from origin = 80/25  = 3.2  

Another method to solve with Equation of lines

perpendicular distance would be shortest

-3i + 4j

4x + 3y = c

putting x= 4 & y = 0

16 + 0 = c

4x + 3y = 16  eq1

perpendicular line would be

-3x + 4y = c

it is at origin

so c = 0

-3x + 4y = 0  eq2

3 × eq1 + 4×eq2

25y = 48

y = 48/25

x = 64/25

distance from origin

= root ( (48/25)^2 + (64/25)^2)

=(16/25) root ( 3^ 2 + 4^2)

= (16/25) × root 25

= (16/25)× 5

= 3.2