A body of mass 3kg moving with a constant acceleration covers a distance of 10m in the 3rd second and lbm in the 4th ssecond respectively.
The initial velocity of the body is

Options

A. `10ms^(-1)`
B. `8ms^(-1)`
C. `5ms^(-1)`
D. `-5ms^(-1)`

Explanation

To solve the problem, we need to find the initial velocity of a body that covers specific distances in the 3rd and 4th seconds of its motion under constant acceleration.

### Step-by-Step Solution:

1. **Understand the formula for distance covered in the nth second:**
The distance covered in the nth second (Sn) can be calculated using the formula:
\[ S_n = u + \frac{a}{2}(2n - 1) \]
where:
- \( S_n \) = distance covered in the nth second
- \( u \) = initial velocity
- \( a \) = acceleration
- \( n \) = the second in which the distance is covered

2. **Set up the equations for the 3rd and 4th seconds:**
- For the 3rd second, we know \( S_3 = 10 \) m:
\[ S_3 = u + \frac{a}{2}(2 \cdot 3 - 1) = u + \frac{a}{2}(5) \]
This gives us:
\[ 10 = u + \frac{5a}{2} \quad \text{(Equation 1)} \]

- For the 4th second, we know \( S_4 = 16 \) m:
\[ S_4 = u + \frac{a}{2}(2 \cdot 4 - 1) = u + \frac{a}{2}(7) \]
This gives us:
\[ 16 = u + \frac{7a}{2} \quad \text{(Equation 2)} \]

3. **Subtract Equation 1 from Equation 2:**
\[ (16 - 10) = \left(u + \frac{7a}{2}\right) - \left(u + \frac{5a}{2}\right) \]
Simplifying this gives:
\[ 6 = \frac{7a}{2} - \frac{5a}{2} \]
\[ 6 = \frac{2a}{2} \implies 6 = a \]
Thus, the acceleration \( a = 6 \, \text{m/s}^2 \).

4. **Substitute the value of acceleration back into Equation 1:**
Now we can substitute \( a \) back into Equation 1 to find \( u \):
\[ 10 = u + \frac{5 \cdot 6}{2} \]
\[ 10 = u + 15 \]
Rearranging gives:
\[ u = 10 - 15 = -5 \, \text{m/s} \]

5. **Conclusion:**
The initial velocity of the body is \( u = -5 \, \text{m/s} \).