A particle is moving along a straight line with constant acceleration. At the end of tenth second its velocity becomes `20 m//s` and tenth second it travels a distance od `10 m`. Then the acceleration of the particle will be.

Options

A. `10 m//s^(2)`
B. `20 m//s^(2)`
C. `(1)/(5) m//s^(2)`
D. `3.8 m//s^(2)`

Explanation

To solve the problem step by step, we will use the equations of motion for a particle moving with constant acceleration.

### Given:
1. Final velocity at the end of the 10th second, \( V = 20 \, \text{m/s} \)
2. Distance traveled during the 10th second, \( S_n = 10 \, \text{m} \)
3. Time, \( t = 10 \, \text{s} \)

### Step 1: Use the first equation of motion
The first equation of motion is given by:
\[ V = U + A t \]
Where:
- \( V \) is the final velocity,
- \( U \) is the initial velocity,
- \( A \) is the acceleration,
- \( t \) is the time.

Substituting the known values:
\[ 20 = U + 10A \quad \text{(Equation 1)} \]

### Step 2: Use the formula for distance traveled in the nth second
The distance traveled in the nth second is given by:
\[ S_n = U + \frac{A}{2}(2n - 1) \]
For the 10th second (\( n = 10 \)):
\[ 10 = U + \frac{A}{2}(2 \cdot 10 - 1) \]
This simplifies to:
\[ 10 = U + \frac{A}{2}(19) \]
Thus,
\[ 10 = U + \frac{19A}{2} \quad \text{(Equation 2)} \]

### Step 3: Solve the equations simultaneously
Now we have two equations:
1. \( 20 = U + 10A \) (Equation 1)
2. \( 10 = U + \frac{19A}{2} \) (Equation 2)

From Equation 1, we can express \( U \) in terms of \( A \):
\[ U = 20 - 10A \]

### Step 4: Substitute \( U \) in Equation 2
Substituting \( U \) from Equation 1 into Equation 2:
\[ 10 = (20 - 10A) + \frac{19A}{2} \]
Now, simplifying this:
\[ 10 = 20 - 10A + \frac{19A}{2} \]
Rearranging gives:
\[ 10 - 20 = -10A + \frac{19A}{2} \]
\[ -10 = -10A + \frac{19A}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ -20 = -20A + 19A \]
\[ -20 = -A \]
Thus, we find:
\[ A = 20 \, \text{m/s}^2 \]

### Conclusion
The acceleration of the particle is \( A = 20 \, \text{m/s}^2 \).

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