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
Answers
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