A body moving with uniform acceleration in a straight line travels a distance of 50 m in `7^(th)` second and a distance of 70 m in `11^(th)` second. How much distance will it cover in 40 seconds?

Options

A. 1400 m
B. 4700 m
C. 3750 m
D. 650 m

Answer

To solve the problem of a body moving with uniform acceleration, we need to find the distance it covers in 40 seconds given that it travels 50 m in the 7th second and 70 m in the 11th second.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We know the distance traveled in the nth second can be given by the formula:
\[ S_n = u + \frac{a}{2}(2n - 1) \]
where \( S_n \) is the distance traveled in the nth second, \( u \) is the initial velocity, \( a \) is the acceleration, and \( n \) is the second.

2. **Setting Up the Equations**:
- For the 7th second:
\[ S_7 = u + \frac{a}{2}(2 \cdot 7 - 1) = u + \frac{a}{2}(13) = 50 \quad \text{(Equation 1)} \]
- For the 11th second:
\[ S_{11} = u + \frac{a}{2}(2 \cdot 11 - 1) = u + \frac{a}{2}(21) = 70 \quad \text{(Equation 2)} \]

3. **Solving the Equations**:
- From Equation 1:
\[ u + \frac{13a}{2} = 50 \quad \text{(1)} \]
- From Equation 2:
\[ u + \frac{21a}{2} = 70 \quad \text{(2)} \]
- Subtract Equation (1) from Equation (2):
\[ \left(u + \frac{21a}{2}\right) - \left(u + \frac{13a}{2}\right) = 70 - 50 \]
\[ \frac{21a}{2} - \frac{13a}{2} = 20 \]
\[ \frac{8a}{2} = 20 \implies 4a = 20 \implies a = 5 \, \text{m/s}^2 \]

4. **Finding Initial Velocity \( u \)**:
- Substitute \( a = 5 \) back into Equation (1):
\[ u + \frac{13 \cdot 5}{2} = 50 \]
\[ u + \frac{65}{2} = 50 \implies u + 32.5 = 50 \]
\[ u = 50 - 32.5 = 17.5 \, \text{m/s} \]

5. **Calculating Distance in 40 seconds**:
- Use the second equation of motion:
\[ S = ut + \frac{1}{2} a t^2 \]
- Substitute \( u = 17.5 \, \text{m/s} \), \( a = 5 \, \text{m/s}^2 \), and \( t = 40 \, \text{s} \):
\[ S = 17.5 \cdot 40 + \frac{1}{2} \cdot 5 \cdot (40^2) \]
\[ S = 700 + \frac{1}{2} \cdot 5 \cdot 1600 \]
\[ S = 700 + \frac{5 \cdot 1600}{2} = 700 + 4000 = 4700 \, \text{m} \]

### Final Answer:
The distance covered in 40 seconds is **4700 meters**.