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 follow the information given in the question and use the equations of motion.

### Step 1: Understand the problem
We have a particle moving with constant acceleration, starting from rest. The distance covered in the last second of its motion is \( \frac{9}{25} \) of the total distance. We also know that the particle covers 6 cm in the first second.

### Step 2: Use the first second distance formula
Since the particle starts from rest, we can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \) (initial velocity), \( s = 6 \) cm (distance in the first second), and \( t = 1 \) s. Thus, we have:
\[ 6 = 0 \cdot 1 + \frac{1}{2} a (1^2) \]
This simplifies to:
\[ 6 = \frac{1}{2} a \]
From which we can find:
\[ a = 12 \text{ cm/s}^2 \]

### Step 3: Set up the equations for total distance
Let \( t \) be the total time of motion. The total distance \( x \) covered by the particle can be expressed as:
\[ x = ut + \frac{1}{2} a t^2 \]
Since \( u = 0 \):
\[ x = \frac{1}{2} a t^2 = \frac{1}{2} \cdot 12 \cdot t^2 = 6t^2 \]

### Step 4: Distance covered in the last second
The distance covered in the last second (i.e., during the \( t \)-th second) can be calculated using the formula:
\[ s_n = u + a \left(n - 1\right) + \frac{1}{2} a \]
For the \( t \)-th second, this becomes:
\[ s_t = 0 + a(t - 1) + \frac{1}{2} a = a(t - 1) + \frac{1}{2} a = a \left(t - \frac{1}{2}\right) \]
Substituting \( a = 12 \):
\[ s_t = 12 \left(t - 1\right) + 6 = 12t - 6 \]

### Step 5: Relate the last second distance to the total distance
According to the problem, the distance covered in the last second is \( \frac{9}{25} \) of the total distance:
\[ 12t - 6 = \frac{9}{25} x \]
Substituting \( x = 6t^2 \):
\[ 12t - 6 = \frac{9}{25} (6t^2) \]
This simplifies to:
\[ 12t - 6 = \frac{54}{25} t^2 \]

### Step 6: Rearranging to form a quadratic equation
Rearranging gives:
\[ \frac{54}{25} t^2 - 12t + 6 = 0 \]
Multiplying through by 25 to eliminate the fraction:
\[ 54t^2 - 300t + 150 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{300 \pm \sqrt{(-300)^2 - 4 \cdot 54 \cdot 150}}{2 \cdot 54} \]
Calculating the discriminant:
\[ b^2 - 4ac = 90000 - 32400 = 57600 \]
Thus:
\[ t = \frac{300 \pm 240}{108} \]
Calculating the two possible values:
1. \( t = \frac{540}{108} = 5 \) seconds
2. \( t = \frac{60}{108} = \frac{5}{9} \) seconds

Since the time must be greater than 1 second, we take \( t = 5 \) seconds.

### Step 8: Calculate the total distance
Now substituting \( t = 5 \) into the equation for total distance:
\[ x = 6t^2 = 6 \cdot (5^2) = 6 \cdot 25 = 150 \text{ cm} \]

### Final Answer
The particle is in motion for **5 seconds** and covers a distance of **150 cm**.

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