Question

Two particles are projected simultaneously towards each other from opposite ends of a straight horizontal tube. One particle travels 97 cm in the 1st 2 seconds, 93 cm in next 2 seconds , 89 cm in another next 2 seconds, etc. The other particle travels 49 cm in the 1st 2 seconds, 47 cm in the next 2 seconds , 45 cm in another next 2 seconds, etc. Find how long it takes for the particles to meet and hence, determine the distances travelled by each particle

Answer 1

The movement of both particles are illustrations of arithmetic progression.

Both particles meet at 1 cm, after 50 seconds

For particle 1, we have:

$\mathbf{Particle\ 1: 97, 93, 89,.....}$

The nth term of particle 1 is:

$\mathbf{T_n = 97 - 4(n - 1)}$

For particle 2, we have:

$\mathbf{Particle\ 2: 49, 47, 45,.....}$

The nth term of particle 2 is:

$\mathbf{T_n = 49 - 2(n - 1)}$

Both particles are at the same distance, when:

$\mathbf{T_n = T_n}$

So, we have:

$\mathbf{97 - 4(n - 1) = 49 - 2(n -1)}$

Subtract 49 from both sides

$\mathbf{48- 4(n - 1) = - 2(n -1)}$

Divide through by -2

$\mathbf{-24+ 2(n - 1) = (n -1)}$

Collect like terms

$\mathbf{-24 = (n -1) - 2(n - 1)}$

$\mathbf{-24 = -(n -1)}$

Divide through by -1

$\mathbf{24 = n -1}$

Add 1 to both sides

$\mathbf{n = 25 }$

Substitute 25 for n in $\mathbf{T_n = 97 - 4(n - 1)}$

$\mathbf{T_{25} = 97 - 4(25 -1)}$

$\mathbf{T_{25} = 1}$

This means that both particles meet at 1 cm

The time they meet is:

$\mathbf{Time = n \times 2\ seconds}$

Substitute 25 for n

$\mathbf{Time = 25 \times 2\ seconds}$

$\mathbf{Time = 50\ seconds}$

Hence, both particles meet after 50 seconds

Read more about arithmetic progressions at:

https://brainly.com/question/13989292