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 information given about the distances covered in specific seconds and the equations of motion for uniformly accelerated motion.

### Step 1: Understand the Problem
We know that a body covers:
- 4 m in the 3rd second
- 12 m in the 5th second

We need to find out how far the body will travel in the next 3 seconds (from the 5th to the 8th second).

### Step 2: Use the Formula for Distance in nth Second
The distance covered in the nth second can be calculated using the formula:
\[ s_n = u + \frac{a}{2} (2n - 1) \]
where:
- \( s_n \) is the distance covered in the nth second,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( n \) is the second number.

### Step 3: Set Up Equations for the Given Seconds
For the 3rd second:
\[ s_3 = 4 = u + \frac{a}{2} (2 \times 3 - 1) \]
This simplifies to:
\[ 4 = u + \frac{5a}{2} \quad \text{(Equation 1)} \]

For the 5th second:
\[ s_5 = 12 = u + \frac{a}{2} (2 \times 5 - 1) \]
This simplifies to:
\[ 12 = u + \frac{9a}{2} \quad \text{(Equation 2)} \]

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

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

### Step 5: Substitute Back to Find Initial Velocity
Now substitute \( a = 4 \) into Equation 1:
\[ 4 = u + \frac{5 \times 4}{2} \]
This simplifies to:
\[ 4 = u + 10 \implies u = 4 - 10 = -6 \, \text{m/s} \]

### Step 6: Calculate Distance Traveled in the Next 3 Seconds
We need to find the distance traveled from the 5th to the 8th second. This can be calculated as:
\[ d = s_8 - s_5 \]
Where:
- \( s_8 \) is the distance traveled in the first 8 seconds,
- \( s_5 \) is the distance traveled in the first 5 seconds.

Using the formula for total distance traveled in time \( t \):
\[ s = ut + \frac{1}{2} a t^2 \]

Calculate \( s_8 \):
\[ s_8 = u \cdot 8 + \frac{1}{2} \cdot a \cdot (8^2) = (-6) \cdot 8 + \frac{1}{2} \cdot 4 \cdot 64 \]
This simplifies to:
\[ s_8 = -48 + 128 = 80 \, \text{m} \]

Calculate \( s_5 \):
\[ s_5 = u \cdot 5 + \frac{1}{2} \cdot a \cdot (5^2) = (-6) \cdot 5 + \frac{1}{2} \cdot 4 \cdot 25 \]
This simplifies to:
\[ s_5 = -30 + 50 = 20 \, \text{m} \]

### Step 7: Find the Distance Traveled in the Next 3 Seconds
Now, calculate \( d \):
\[ d = s_8 - s_5 = 80 - 20 = 60 \, \text{m} \]

### Final Answer
The body will travel **60 meters** in the next 3 seconds.