A body starts from rest with an acceleration `2m//s^(2)` till it attains the maximum velocity then retards to rest with `3m//s^(2)`. If total time taken is 10 second, then maximum speed attained is

Options

A. 12 m/s
B. 8 m/s
C. 6 m/s
D. 4 m/s

Answer

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

### Step 1: Understand the motion phases
The body starts from rest, accelerates with an acceleration of \(2 \, \text{m/s}^2\) until it reaches its maximum velocity, and then retards to rest with a deceleration of \(3 \, \text{m/s}^2\). The total time for both phases is given as \(10\) seconds.

### Step 2: Define variables
Let:
- \(V_{\text{max}}\) = maximum velocity attained
- \(t_1\) = time taken to reach maximum velocity
- \(t_2\) = time taken to come to rest

From the problem, we know:
\[ t_1 + t_2 = 10 \, \text{seconds} \]

### Step 3: Use kinematic equations for the acceleration phase
During the acceleration phase, the body starts from rest, so:
- Initial velocity \(u = 0\)
- Acceleration \(a_1 = 2 \, \text{m/s}^2\)

Using the kinematic equation:
\[ V_{\text{max}} = u + a_1 t_1 \]
Substituting the known values:
\[ V_{\text{max}} = 0 + 2 t_1 \]
Thus, we have:
\[ V_{\text{max}} = 2 t_1 \quad \text{(Equation 1)} \]

### Step 4: Use kinematic equations for the deceleration phase
During the deceleration phase:
- Initial velocity \(u = V_{\text{max}}\)
- Final velocity \(v = 0\)
- Deceleration \(a_2 = -3 \, \text{m/s}^2\)

Using the kinematic equation:
\[ v = u + a_2 t_2 \]
Substituting the known values:
\[ 0 = V_{\text{max}} - 3 t_2 \]
Rearranging gives:
\[ V_{\text{max}} = 3 t_2 \quad \text{(Equation 2)} \]

### Step 5: Relate \(t_1\) and \(t_2\) using total time
From the total time equation:
\[ t_1 + t_2 = 10 \]
We can express \(t_2\) in terms of \(t_1\):
\[ t_2 = 10 - t_1 \quad \text{(Equation 3)} \]

### Step 6: Substitute \(t_2\) into Equation 2
Substituting Equation 3 into Equation 2:
\[ V_{\text{max}} = 3(10 - t_1) \]
Thus:
\[ V_{\text{max}} = 30 - 3t_1 \quad \text{(Equation 4)} \]

### Step 7: Set Equations 1 and 4 equal to each other
Now we have two expressions for \(V_{\text{max}}\):
1. \(V_{\text{max}} = 2 t_1\)
2. \(V_{\text{max}} = 30 - 3t_1\)

Setting them equal:
\[ 2 t_1 = 30 - 3 t_1 \]

### Step 8: Solve for \(t_1\)
Rearranging gives:
\[ 2 t_1 + 3 t_1 = 30 \]
\[ 5 t_1 = 30 \]
\[ t_1 = 6 \, \text{seconds} \]

### Step 9: Find \(t_2\)
Using Equation 3:
\[ t_2 = 10 - t_1 = 10 - 6 = 4 \, \text{seconds} \]

### Step 10: Calculate \(V_{\text{max}}\)
Using Equation 1:
\[ V_{\text{max}} = 2 t_1 = 2 \times 6 = 12 \, \text{m/s} \]

### Final Answer
The maximum speed attained by the body is \(12 \, \text{m/s}\).

---