A body moving with uniform acceleration covers a distance of 14 m in first 2 second and 88 m in next 4 second. How much distance is travelled by it in `5^(th)` second?

Answer

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

### Step 1: Understand the Problem
We know that a body is moving with uniform acceleration. It covers:
- 14 m in the first 2 seconds
- 88 m in the next 4 seconds (from 2 seconds to 6 seconds)

We need to find the distance covered in the 5th second.

### Step 2: Use the Equation of Motion
The equation of motion we will use is:
\[ S = Ut + \frac{1}{2}At^2 \]
where:
- \( S \) = distance covered
- \( U \) = initial velocity
- \( A \) = acceleration
- \( t \) = time

### Step 3: Set Up Equations for the First Interval (0 to 2 seconds)
For the first 2 seconds:
\[ S_1 = 14 \, \text{m} \]
\[ t_1 = 2 \, \text{s} \]

Using the equation:
\[ 14 = U(2) + \frac{1}{2}A(2^2) \]
This simplifies to:
\[ 14 = 2U + 2A \]
Dividing through by 2 gives us:
\[ 7 = U + A \quad \text{(Equation 1)} \]

### Step 4: Set Up Equations for the Second Interval (2 to 6 seconds)
For the next 4 seconds (from 2 seconds to 6 seconds):
\[ S_2 = 88 \, \text{m} \]
\[ t_2 = 4 \, \text{s} \]

Using the equation:
\[ 88 = U(4) + \frac{1}{2}A(4^2) \]
This simplifies to:
\[ 88 = 4U + 8A \]
Dividing through by 4 gives us:
\[ 22 = U + 2A \quad \text{(Equation 2)} \]

### Step 5: Solve the Equations Simultaneously
Now we have two equations:
1. \( U + A = 7 \)
2. \( U + 2A = 22 \)

Subtract Equation 1 from Equation 2:
\[ (U + 2A) - (U + A) = 22 - 7 \]
This simplifies to:
\[ A = 15 \]

### Step 6: Substitute Back to Find U
Substituting \( A = 15 \) back into Equation 1:
\[ U + 15 = 7 \]
Thus:
\[ U = 7 - 15 = -8 \]

### Step 7: Calculate Distance in the 5th Second
To find the distance covered in the 5th second, we can use the formula:
\[ S_n = U + A(2n - 1) \]
where \( n \) is the second we are interested in (5 in this case):
\[ S_5 = U + A(2 \cdot 5 - 1) \]
Substituting the values:
\[ S_5 = -8 + 15(9) \]
\[ S_5 = -8 + 135 \]
\[ S_5 = 127 \]

### Final Answer
The distance travelled by the body in the 5th second is **127 meters**.

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