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

Options

A. 5 s, 150 cm
B. 10 s, 150 cm
C. 15 s, 100 cm
D. None

Answer

To solve the problem step by step, we will use the equations of motion under constant acceleration.

### Step 1: Understand the given information
- The particle starts from rest, so the initial velocity \( u = 0 \).
- The distance covered in the last second of motion is \( \frac{9}{25} \) of the total distance.
- The distance covered in the first second is \( 6 \, \text{cm} \).

### Step 2: Set up the equations
Let \( t_0 \) be the total time of motion, and \( S \) be the total distance covered.

From the equations of motion, the total distance \( S \) covered in \( t_0 \) seconds is given by:
\[ S = ut_0 + \frac{1}{2} a t_0^2 = \frac{1}{2} a t_0^2 \quad \text{(since } u = 0\text{)} \]

The distance covered in the last second can be expressed using the formula for distance covered in the \( n \)-th second:
\[ S_n = u + \frac{1}{2} a (2n - 1) = \frac{1}{2} a (2t_0 - 1) \quad \text{(for } n = t_0\text{)} \]

### Step 3: Relate the last second distance to the total distance
According to the problem, the distance covered in the last second is:
\[ \frac{1}{2} a (2t_0 - 1) = \frac{9}{25} S \]
Substituting \( S \):
\[ \frac{1}{2} a (2t_0 - 1) = \frac{9}{25} \left(\frac{1}{2} a t_0^2\right) \]
Cancelling \( \frac{1}{2} a \) from both sides (assuming \( a \neq 0 \)):
\[ 2t_0 - 1 = \frac{9}{25} t_0^2 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 9t_0^2 - 50t_0 + 25 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( t_0 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 9, b = -50, c = 25 \):
\[ t_0 = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 9 \cdot 25}}{2 \cdot 9} \]
Calculating the discriminant:
\[ t_0 = \frac{50 \pm \sqrt{2500 - 900}}{18} = \frac{50 \pm \sqrt{1600}}{18} = \frac{50 \pm 40}{18} \]
Calculating the two possible values:
1. \( t_0 = \frac{90}{18} = 5 \)
2. \( t_0 = \frac{10}{18} = \frac{5}{9} \) (not physically meaningful since time cannot be negative)

Thus, \( t_0 = 5 \, \text{s} \).

### Step 6: Calculate the acceleration
Using the distance covered in the first second:
\[ S_1 = u + \frac{1}{2} a (2 \cdot 1 - 1) = \frac{1}{2} a = 6 \, \text{cm} \]
Thus, \( a = 12 \, \text{cm/s}^2 \).

### Step 7: Calculate the total distance
Now, substituting \( a \) and \( t_0 \) into the total distance formula:
\[ S = \frac{1}{2} a t_0^2 = \frac{1}{2} \cdot 12 \cdot 5^2 = \frac{1}{2} \cdot 12 \cdot 25 = 150 \, \text{cm} \]

### Final Answer
- The time of motion \( t_0 = 5 \, \text{s} \)
- The total distance covered \( S = 150 \, \text{cm} \)