A body covers a distance of 4 m in `3^(rd)` second and 12m in `5^(th)` second. If the motion is uniformly accelerated. How far will it travel in the next 3 seconds?

Options

A. 10 m
B. 30 m
C. 40 m
D. 60 m

Explanation

To solve the problem step by step, we will use the equations of motion for uniformly accelerated motion.

### Step 1: Understand the given information
We know that:
- The distance covered in the 3rd second (S3) = 4 m
- The distance covered in the 5th second (S5) = 12 m

### Step 2: Use the formula for distance covered in the nth second
The distance covered in the nth second can be expressed as:
\[ S_n = u + \frac{1}{2} a (2n - 1) \]
where:
- \( S_n \) = distance covered in the nth second
- \( u \) = initial velocity
- \( a \) = acceleration
- \( n \) = the second in which we are measuring the distance

### Step 3: Set up equations for S3 and S5
Using the formula for \( S_3 \) and \( S_5 \):
1. For \( n = 3 \):
\[ S_3 = u + \frac{1}{2} a (2 \cdot 3 - 1) = 4 \]
\[ u + \frac{1}{2} a (5) = 4 \]
\[ u + \frac{5}{2} a = 4 \quad \text{(Equation 1)} \]

2. For \( n = 5 \):
\[ S_5 = u + \frac{1}{2} a (2 \cdot 5 - 1) = 12 \]
\[ u + \frac{1}{2} a (9) = 12 \]
\[ u + \frac{9}{2} a = 12 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
Now, we have two equations:
1. \( u + \frac{5}{2} a = 4 \) (Equation 1)
2. \( u + \frac{9}{2} a = 12 \) (Equation 2)

Subtract Equation 1 from Equation 2:
\[ \left( u + \frac{9}{2} a \right) - \left( u + \frac{5}{2} a \right) = 12 - 4 \]
\[ \frac{9}{2} a - \frac{5}{2} a = 8 \]
\[ \frac{4}{2} a = 8 \implies 2a = 8 \implies a = 4 \, \text{m/s}^2 \]

Now substitute \( a \) back into Equation 1 to find \( u \):
\[ u + \frac{5}{2} \cdot 4 = 4 \]
\[ u + 10 = 4 \implies u = 4 - 10 = -6 \, \text{m/s} \]

### Step 5: Find the distance traveled in the next 3 seconds
We need to find the distance covered from the 6th to the 8th second, which is \( S_6 + S_7 + S_8 \).

Using the formula for each second:
1. For \( n = 6 \):
\[ S_6 = -6 + \frac{1}{2} \cdot 4 (2 \cdot 6 - 1) = -6 + 4 \cdot 11 = -6 + 44 = 38 \, \text{m} \]

2. For \( n = 7 \):
\[ S_7 = -6 + \frac{1}{2} \cdot 4 (2 \cdot 7 - 1) = -6 + 4 \cdot 13 = -6 + 52 = 46 \, \text{m} \]

3. For \( n = 8 \):
\[ S_8 = -6 + \frac{1}{2} \cdot 4 (2 \cdot 8 - 1) = -6 + 4 \cdot 15 = -6 + 60 = 54 \, \text{m} \]

### Step 6: Total distance in the next 3 seconds
The total distance traveled in the next 3 seconds is:
\[ S_6 + S_7 + S_8 = 38 + 46 + 54 = 138 \, \text{m} \]

### Final Answer
The body will travel **138 meters** in the next 3 seconds.

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