The velocity of a particle at an instant is `10 m s^(-1)`. After 3 s its velocity will becomes `16 m s^(-1)`. The velocity at 2 s, before the given instant will be

Options

A. `6 m s^(-1)`
B. `4 m s^(-1)`
C. `2 m s^(-1)`
D. `1 m s^(-1)`

Explanation

To solve the problem step by step, we will use the first equation of motion, which relates initial velocity, final velocity, acceleration, and time.

### Step 1: Identify the known values
- Initial velocity at point B (v_B) = 10 m/s (at the given instant)
- Final velocity at point C (v_C) = 16 m/s (after 3 seconds)
- Time interval (t) = 3 s

### Step 2: Use the first equation of motion from point B to point C
The first equation of motion is given by:
\[ v = u + at \]
Where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

Applying this equation from point B to point C:
\[ v_C = v_B + a \cdot t \]
Substituting the known values:
\[ 16 = 10 + a \cdot 3 \]

### Step 3: Solve for acceleration (a)
Rearranging the equation to find acceleration:
\[ 16 - 10 = a \cdot 3 \]
\[ 6 = 3a \]
\[ a = \frac{6}{3} = 2 \, \text{m/s}^2 \]

### Step 4: Now, find the velocity at point A (2 seconds before point B)
We need to find the velocity at point A (v_A) which is 2 seconds before point B. We will again use the first equation of motion, this time from point A to point B.

Using the same equation:
\[ v_B = v_A + a \cdot t \]
Where:
- \( v_B = 10 \, \text{m/s} \)
- \( a = 2 \, \text{m/s}^2 \)
- \( t = 2 \, \text{s} \)

Substituting the values:
\[ 10 = v_A + 2 \cdot 2 \]

### Step 5: Solve for v_A
Rearranging the equation:
\[ 10 = v_A + 4 \]
\[ v_A = 10 - 4 = 6 \, \text{m/s} \]

### Final Answer
The velocity at 2 seconds before the given instant is **6 m/s**.

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