A particle starting from rest has a constant acceleration of `4 m//s^2` for 4 s. It then retards uniformly for next 8 s and comes to rest. Find during the motion of particle (a) average acceleration (b) average speed and (c) average velocity.

Explanation

To solve the problem step by step, we will break down the motion of the particle into two parts: the acceleration phase and the deceleration phase.

### Step 1: Calculate the final velocity after the acceleration phase

The particle starts from rest and has a constant acceleration of \(4 \, \text{m/s}^2\) for \(4 \, \text{s}\).

Using the equation of motion:
\[ v = u + at \]
where:
- \(u = 0 \, \text{m/s}\) (initial velocity),
- \(a = 4 \, \text{m/s}^2\) (acceleration),
- \(t = 4 \, \text{s}\) (time).

Substituting the values:
\[ v = 0 + (4)(4) = 16 \, \text{m/s} \]

### Step 2: Calculate the distance covered during the acceleration phase

Using the equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the values:
\[ s = 0 \cdot 4 + \frac{1}{2} \cdot 4 \cdot (4^2) = \frac{1}{2} \cdot 4 \cdot 16 = 32 \, \text{m} \]

### Step 3: Calculate the retardation during the deceleration phase

The particle comes to rest in \(8 \, \text{s}\) after reaching \(16 \, \text{m/s}\).

Using the equation:
\[ v = u + at \]
where:
- \(v = 0 \, \text{m/s}\) (final velocity),
- \(u = 16 \, \text{m/s}\) (initial velocity),
- \(t = 8 \, \text{s}\).

Rearranging for \(a\):
\[ 0 = 16 + a(8) \implies a = -\frac{16}{8} = -2 \, \text{m/s}^2 \]

### Step 4: Calculate the distance covered during the deceleration phase

Using the same equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substituting the values for the deceleration phase:
\[ s = 16 \cdot 8 + \frac{1}{2} \cdot (-2) \cdot (8^2) \]
Calculating:
\[ s = 128 - \frac{1}{2} \cdot 2 \cdot 64 = 128 - 64 = 64 \, \text{m} \]

### Step 5: Calculate total distance and total time

Total distance \(S\) is the sum of distances from both phases:
\[ S = 32 + 64 = 96 \, \text{m} \]

Total time \(T\) is:
\[ T = 4 + 8 = 12 \, \text{s} \]

### Step 6: Calculate average acceleration

Average acceleration \(a_{avg}\) is given by:
\[ a_{avg} = \frac{v_f - v_i}{T} \]
where:
- \(v_f = 0 \, \text{m/s}\) (final velocity),
- \(v_i = 0 \, \text{m/s}\) (initial velocity).

Substituting the values:
\[ a_{avg} = \frac{0 - 0}{12} = 0 \, \text{m/s}^2 \]

### Step 7: Calculate average speed and average velocity

Average speed \(v_{avg}\) is given by:
\[ v_{avg} = \frac{S}{T} = \frac{96}{12} = 8 \, \text{m/s} \]

Since the total displacement and distance are the same, the average velocity is also:
\[ v_{avg} = 8 \, \text{m/s} \]

### Final Answers:
(a) Average acceleration: \(0 \, \text{m/s}^2\)
(b) Average speed: \(8 \, \text{m/s}\)
(c) Average velocity: \(8 \, \text{m/s}\)