A body covers `4.9m, 14.7 m, 24.5 m and 34.3m` in the successive seconds of motion. The acceleration of the body is

Options

A. `2.7 ms ^(-2)`
B. `4.9 ms ^(-2)`
C. `9.8 ms ^(-2)`
D. `19.6 ms ^(-2)`

Answer

To find the acceleration of the body given the distances covered in successive seconds, we can use the equations of motion. Here’s how we can solve the problem step by step:

### Step 1: Understand the distances covered
The body covers the following distances in successive seconds:
- 1st second: \( s_1 = 4.9 \, \text{m} \)
- 2nd second: \( s_2 = 14.7 \, \text{m} \)
- 3rd second: \( s_3 = 24.5 \, \text{m} \)
- 4th second: \( s_4 = 34.3 \, \text{m} \)

### Step 2: Calculate the total distance covered
The total distance covered in 4 seconds is:
\[ S = s_1 + s_2 + s_3 + s_4 = 4.9 + 14.7 + 24.5 + 34.3 = 78.4 \, \text{m} \]

### Step 3: Use the equation of motion
We can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
For the first second, we can write:
\[ s_1 = u \cdot 1 + \frac{1}{2} a \cdot 1^2 \]
Substituting \( s_1 = 4.9 \):
\[ 4.9 = u + \frac{1}{2} a \quad \text{(Equation 1)} \]

### Step 4: Write the equation for total distance
For the total distance covered in 4 seconds:
\[ S = u \cdot 4 + \frac{1}{2} a \cdot 4^2 \]
Substituting \( S = 78.4 \):
\[ 78.4 = 4u + \frac{1}{2} a \cdot 16 \]
This simplifies to:
\[ 78.4 = 4u + 8a \quad \text{(Equation 2)} \]

### Step 5: Solve the equations
Now we have two equations:
1. \( 4.9 = u + \frac{1}{2} a \) (Equation 1)
2. \( 78.4 = 4u + 8a \) (Equation 2)

From Equation 1, we can express \( u \):
\[ u = 4.9 - \frac{1}{2} a \]

Substituting this expression for \( u \) into Equation 2:
\[ 78.4 = 4(4.9 - \frac{1}{2} a) + 8a \]
Expanding this gives:
\[ 78.4 = 19.6 - 2a + 8a \]
Combining like terms:
\[ 78.4 = 19.6 + 6a \]
Subtracting \( 19.6 \) from both sides:
\[ 58.8 = 6a \]
Now, dividing by 6:
\[ a = \frac{58.8}{6} = 9.8 \, \text{m/s}^2 \]

### Conclusion
The acceleration of the body is \( 9.8 \, \text{m/s}^2 \).

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