A particle moving with uniform retardation covers distances 18 m, 14 m and 10 m in successive seconds. It comes to rest after travelling a further distance of

Options

A. 50 m
B. 8 m
C. 12 m
D. 42 m

Explanation

To solve the problem step by step, we will analyze the motion of the particle and use the equations of motion to find the required distance it travels before coming to rest.

### Step 1: Understand the Given Information
The particle covers distances of:
- \( s_1 = 18 \, \text{m} \) in the first second,
- \( s_2 = 14 \, \text{m} \) in the second second,
- \( s_3 = 10 \, \text{m} \) in the third second.

### Step 2: Use the Equation of Motion
The distance covered in the nth second is given by:
\[ s_n = u + \frac{a}{2}(2n - 1) \]
where \( u \) is the initial velocity, \( a \) is the acceleration (negative for retardation), and \( n \) is the second.

### Step 3: Set Up Equations
For the first second (\( n = 1 \)):
\[ s_1 = u + \frac{a}{2}(2 \cdot 1 - 1) = u + \frac{a}{2} = 18 \quad \text{(1)} \]

For the second second (\( n = 2 \)):
\[ s_2 = u + \frac{a}{2}(2 \cdot 2 - 1) = u + \frac{3a}{2} = 14 \quad \text{(2)} \]

### Step 4: Solve the Equations
From equation (1):
\[ u + \frac{a}{2} = 18 \implies u = 18 - \frac{a}{2} \quad \text{(3)} \]

Substituting (3) into equation (2):
\[ (18 - \frac{a}{2}) + \frac{3a}{2} = 14 \]
Simplifying:
\[ 18 + a = 14 \implies a = 14 - 18 = -4 \, \text{m/s}^2 \]

### Step 5: Find Initial Velocity
Substituting \( a = -4 \) into equation (3):
\[ u = 18 - \frac{-4}{2} = 18 + 2 = 20 \, \text{m/s} \]

### Step 6: Calculate Distance to Come to Rest
Using the equation:
\[ v^2 = u^2 + 2aS \]
where \( v = 0 \) (final velocity), \( u = 20 \, \text{m/s} \), and \( a = -4 \, \text{m/s}^2 \):
\[ 0 = (20)^2 + 2(-4)S \]
\[ 0 = 400 - 8S \]
\[ 8S = 400 \implies S = \frac{400}{8} = 50 \, \text{m} \]

### Step 7: Total Distance Covered
The total distance covered before coming to rest is:
\[ s_{\text{total}} = s_1 + s_2 + s_3 + S = 18 + 14 + 10 + 50 = 92 \, \text{m} \]

### Step 8: Distance After Last Recorded Distance
The distance covered before coming to rest after the last recorded distance is:
\[ \text{Distance after last recorded} = S - (s_1 + s_2 + s_3) = 50 - (18 + 14 + 10) = 50 - 42 = 8 \, \text{m} \]

### Final Answer
The particle comes to rest after traveling a further distance of **8 meters**.

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