A particle moves along the x-direction with constant acceleration. The displacement, measured from a convenient position, is 2 m at time `t= 0` and is zero when `t= 10 s.` If the velocity of the particle is momentary zero when `t = 6 s,` determine the acceleration a and the velocity v when `t= 10s.`

Explanation

To solve the problem step by step, we will use the equations of motion under constant acceleration.

### Step 1: Identify the known values
- Initial displacement \( s_0 = 2 \, \text{m} \) at \( t = 0 \, \text{s} \)
- Displacement \( s = 0 \, \text{m} \) at \( t = 10 \, \text{s} \)
- Velocity \( v = 0 \) at \( t = 6 \, \text{s} \)

### Step 2: Use the equation of motion for displacement
The equation of motion for displacement is given by:
\[ s = s_0 + ut + \frac{1}{2} a t^2 \]
At \( t = 0 \):
\[ s_0 = 2 \, \text{m} \]
At \( t = 10 \, \text{s} \):
\[ 0 = 2 + 10u + \frac{1}{2} a (10^2) \]
This simplifies to:
\[ 0 = 2 + 10u + 50a \quad \text{(1)} \]

### Step 3: Use the equation for velocity
The equation for velocity is:
\[ v = u + at \]
At \( t = 6 \, \text{s} \):
\[ 0 = u + 6a \quad \text{(2)} \]

### Step 4: Solve the equations simultaneously
From equation (2), we can express \( u \) in terms of \( a \):
\[ u = -6a \]
Now substitute \( u \) in equation (1):
\[ 0 = 2 + 10(-6a) + 50a \]
This simplifies to:
\[ 0 = 2 - 60a + 50a \]
\[ 0 = 2 - 10a \]
\[ 10a = 2 \implies a = 0.2 \, \text{m/s}^2 \]

### Step 5: Find the initial velocity \( u \)
Substituting \( a \) back into equation (2):
\[ u = -6(0.2) = -1.2 \, \text{m/s} \]

### Step 6: Find the velocity at \( t = 10 \, \text{s} \)
Using the velocity equation again:
\[ v = u + at \]
Substituting \( u \) and \( a \):
\[ v = -1.2 + 0.2 \times 10 \]
\[ v = -1.2 + 2 = 0.8 \, \text{m/s} \]

### Final Results
- Acceleration \( a = 0.2 \, \text{m/s}^2 \)
- Velocity at \( t = 10 \, \text{s} \) is \( v = 0.8 \, \text{m/s} \)