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?

options
1. 5s,150 cm
2. 10s,150 cm

Assume the particle's acceleration is a cm/sec^2 and that it moves a distance of S(t)cms in total time of t secs. Also it covers S(t-1) cms. in (t-1) secs.

Since it covers 6cm in the first second => 6 = 0.1 + (1/2).a.1^2 => a=12cm/sec^2.

Now, S(t)-S(t-1)= (1/2).a.[t^2-(t-1)^2]......(1)

Also, S(t)-S(t-1)= (9/25).S(t)= (9/25).(1/2).a.t^2 .....(2)
From the above two equations, you will get the following eqn.
9t^2-50t+25 = 0
Solve it to get two values for t i.e. t=5,10/18 (ignore the second as t is not less than 1sec)
Thus t=5sec and hence S(t)=150cm
Therefore,option (1) is correct.

=

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

\[ S = ut + \frac{1}{2}at^2 \]

Where:
- S is the distance traveled,
- u is the initial velocity (which is zero since it starts from rest),
- a is the constant acceleration, and
- t is the time taken.

Given that the particle describes 6cm in the first second, we can substitute the known values into the equation and solve for a:

\[ 6 = 0 + \frac{1}{2}a(1^2) \]
\[ 6 = \frac{1}{2}a \]
\[ a = 12 \]

Now, let's calculate the time it takes for the particle to describe 9/25th of the total distance:

\[ S = ut + \frac{1}{2}at^2 \]

Since the particle started from rest, the equation simplifies to:

\[ S = \frac{1}{2}at^2 \]

Given that the particle describes 9/25th of the total distance in the last second, let's assume the total time the particle is in motion is "T" seconds. Therefore, the time taken to describe 9/25th of the distance would be T - 1 sec. We can now substitute known values again and solve for T:

\[ \frac{9}{25}S = \frac{1}{2}a(T-1)^2 \]

Substituting the value of a calculated earlier:

\[ \frac{9}{25}S = \frac{1}{2}(12)(T-1)^2 \]
\[ \frac{9}{25}S = 6(T-1)^2 \]

Given that the particle traveled 6cm in the first second, we can substitute S = 6cm:

\[ \frac{9}{25}(6) = 6(T-1)^2 \]
\[ \frac{54}{25} = 6(T-1)^2 \]
\[ (T-1)^2 = \frac{54}{25} \]
\[ T-1 = \sqrt{\frac{54}{25}} \]
\[ T-1 = \frac{\sqrt{54}}{\sqrt{25}} \]
\[ T-1 = \frac{3\sqrt{6}}{5} \]
\[ T = \frac{3\sqrt{6}}{5} + 1 \]

Now we know the total time the particle is in motion (T) and the total distance traveled (S). We can calculate the distance traveled using the equation:

\[ S = \frac{1}{2}at^2 \]

Substituting known values:

\[ S = \frac{1}{2}(12)(T-1)^2 \]

\[ S = \frac{1}{2}(12)[\frac{3\sqrt{6}}{5} + 1 - 1]^2 \]
\[ S = \frac{1}{2}(12)(\frac{3\sqrt{6}}{5})^2 \]
\[ S = \frac{3\sqrt{6}}{5} \times 3\sqrt{6} \]
\[ S = \frac{9}{5} \times 6 \]
\[ S = \frac{54}{5} \]
\[ S = 10.8 \text{ cm} \]

Therefore, the particle is in motion for \boxed{10} seconds and travels a distance of \boxed{10.8} cm.

To solve this problem, we need to use the equations of motion. The equation that relates the distance traveled, initial velocity, time, and acceleration is given by:

s = ut + 0.5at^2

Where:
- s is the total distance traveled
- u is the initial velocity (which is zero since the particle starts from rest)
- a is the constant acceleration
- t is the time interval of motion

Let's break down the information given in the question:

1. The particle describes 6 cm in the first second.
From this, we can determine the initial distance (s1) traveled in the first second using the equation:

6 cm = 0 + 0.5a(1)^2

Since we know the distance traveled and the time, we can solve for the acceleration (a):

6 = 0.5a

a = 12 cm/s^2

2. In the last second of its motion, the particle travels 9/25th of the whole distance.
Let's assume the total time of motion is t. From this information, we can determine the remaining distance (s2) traveled in the remaining time (t - 1 second):

s2 = (9/25)(s - 6)

Now, we have two equations:

s1 = 6 cm
s2 = (9/25)(s - 6)

We want to find the total distance traveled (s) and the total time of motion (t).

Solving for s1, we find that a = 12 cm/s^2.

For s2, we know that it is equal to (9/25) times the remaining distance (s - 6). So, s2 = (9/25)(s - 6).

Therefore, s2 = (9/25)(s - 6) = 9/25s - 54/25.

To find t, we need to find the time it takes to travel the remaining distance, which is t - 1 second.

From the equation of motion for s2 (s2 = (9/25)s - 54/25), we can plug in t - 1 second for t:

(9/25)(s - 6) = (9/25)(t - 1)^2

To solve for t, we need to solve this equation.

Now, we have two equations:
s1 = 6 cm
(9/25)(s - 6) = (9/25)(t - 1)^2

By solving these equations simultaneously, we can find the values of s and t.

Unfortunately, the question does not provide enough information to determine the exact values of s and t. So, we cannot determine the correct option from the given choices (1. 5s, 150 cm or 2. 10s, 150 cm).