A particle moving with a constant acceleration describes in the last second of its motion 9/25th of its total distance. If it starts from rest, how long is the particle in motion and through what distance does it move if it describes 6cm in the first sec.?





The problem describes a situation where a particle is moving with a constant acceleration. We are given that the particle starts from rest, and in the last second of its motion, it covers 9/25th of its total distance. We are also given that the particle covers 6cm in the first second of its motion. We are required to find the total time for which the particle is in motion and the distance it covers in that time.



To solve this problem, we can use the kinematic equations of motion. Let us assume that the particle covers a total distance of S in time t and that its acceleration is a. Then we can use the following equations:

- S = ut + 1/2 at^2 (Equation 1)
- v = u + at (Equation 2)
- v^2 = u^2 + 2as (Equation 3)

where u is the initial velocity (which is zero in this case), v is the final velocity, and s is the displacement.



Let us first calculate the total time for which the particle is in motion. We know that in the last second of its motion, the particle covers 9/25th of its total distance. Therefore, in the remaining time, it covers 16/25th of the total distance. Let us assume that the total time for which the particle is in motion is T. Then we can write:

- Distance covered in the last second = 9/25 * S
- Distance covered in the remaining time = 16/25 * S
- Distance covered in the first second = 6cm

Therefore, we can write:

- 9/25 * S = S - 16/25 * S
- 9/25 * S + 16/25 * S = S
- S = 25/9 * (6 + T - 1)^2 - 6 (Using Equation 1)

We can simplify this equation to get:

- 25/9 * (T - 5)^2 = S/9

Now, we know that the particle starts from rest and that it has a constant acceleration. Therefore, we can use Equation 2 to find the final velocity of the particle after the first second:

- v = u + at
- v = 0 + a * 1
- v = a



We can use Equation 3 to find the distance covered by the particle in time T:

- v^2 = u^2 + 2as
- a^2 = 2as (since u = 0)
- s = a^2 / (2a)
- s = a / 2

Now, we know