The movement of both particles are illustrations of arithmetic progression.
Both particles meet at 1 cm, after 50 seconds
For particle 1, we have:
[tex]\mathbf{Particle\ 1: 97, 93, 89,.....}[/tex]
The nth term of particle 1 is:
[tex]\mathbf{T_n = 97 - 4(n - 1)}[/tex]
For particle 2, we have:
[tex]\mathbf{Particle\ 2: 49, 47, 45,.....}[/tex]
The nth term of particle 2 is:
[tex]\mathbf{T_n = 49 - 2(n - 1)}[/tex]
Both particles are at the same distance, when:
[tex]\mathbf{T_n = T_n}[/tex]
So, we have:
[tex]\mathbf{97 - 4(n - 1) = 49 - 2(n -1)}[/tex]
Subtract 49 from both sides
[tex]\mathbf{48- 4(n - 1) = - 2(n -1)}[/tex]
Divide through by -2
[tex]\mathbf{-24+ 2(n - 1) = (n -1)}[/tex]
Collect like terms
[tex]\mathbf{-24 = (n -1) - 2(n - 1)}[/tex]
[tex]\mathbf{-24 = -(n -1)}[/tex]
Divide through by -1
[tex]\mathbf{24 = n -1}[/tex]
Add 1 to both sides
[tex]\mathbf{n = 25 }[/tex]
Substitute 25 for n in [tex]\mathbf{T_n = 97 - 4(n - 1)}[/tex]
[tex]\mathbf{T_{25} = 97 - 4(25 -1)}[/tex]
[tex]\mathbf{T_{25} = 1}[/tex]
This means that both particles meet at 1 cm
The time they meet is:
[tex]\mathbf{Time = n \times 2\ seconds}[/tex]
Substitute 25 for n
[tex]\mathbf{Time = 25 \times 2\ seconds}[/tex]
[tex]\mathbf{Time = 50\ seconds}[/tex]
Hence, both particles meet after 50 seconds
Read more about arithmetic progressions at:
https://brainly.com/question/13989292
