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

In its first t-1 seconds, the particle moves 16/25 of the distance covered in t seconds.


(a/2)(t-1)^2 = (16/25)(a/2)t^2
(t-1)^2/t^2 = 16/25
(t-1)/t = 3/4
t = 4 seconds

(a/2)*1^2 = 0.06 m
a = 0.12 m/s^2

In 4 seconds, it moves
(a/2)*4^2 = 0.06*16 = 0.96 m

To solve this problem, we can use the equations of motion for an object moving with constant acceleration.

Let's assign some variables to the given information:
- Let "a" be the constant acceleration of the particle.
- Let "t" represent the time for which the particle is in motion.
- Let "d" represent the total distance traveled by the particle.

We are given that the particle describes 6 cm (0.06 m) in the first second of its motion. We can use this information to find the acceleration:

Using the first equation of motion: d = ut + (1/2)at^2
Substituting the values: 0.06 = 0*t + (1/2)*a*(1^2)
Simplifying the equation: 0.06 = (1/2)*a

Now, we are also given that in the last second of its motion, the particle covers 9/25th of the whole distance. Let's calculate that distance:

Using the same equation of motion: d = ut + (1/2)at^2
Substituting the values: d = 0*t + (1/2)*a*(2t)^2
Simplifying the equation: d = 2ta^2

We know that in the last second, which is t = 1, the particle covers 9/25th of the whole distance:
(9/25)*d = 2*a*(1^2)

Now, we have two equations:
1) 0.06 = (1/2)*a
2) (9/25)*d = 2*a

To find the value of "a," we can solve equation (1):
0.06 = (1/2)*a
a = 0.12 m/s^2

Now, we can substitute this value of "a" into equation (2):
(9/25)*d = 2*(0.12)*(1^2)
(9/25)*d = 0.24

Solving for "d":
(9/25)*d = 0.24
d = (0.24*25) / 9
d = 0.6667 m or approximately 0.67 m

Therefore, the particle is in motion for 2 seconds (first second + last second) and travels a distance of approximately 0.67 meters.