A particle moves for 8s. It first accelerates from rest and then retards to rest. If the re tardation be 3 times the acceleration, then 1 time for which it accelerates will be

Options

A. 2s
B. 3s
C. 4s
D. 6s

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Define Variables
Let:
- \( t_1 \) = time for which the particle accelerates
- \( t_2 \) = time for which the particle decelerates
- \( a \) = acceleration
- \( d \) = retardation (deceleration)

Given that the retardation \( d \) is 3 times the acceleration \( a \):
\[ d = 3a \]

### Step 2: Write the Total Time Equation
The total time of motion is given as 8 seconds:
\[ t_1 + t_2 = 8 \]

### Step 3: Relate Acceleration and Deceleration Times
Since the particle accelerates for \( t_1 \) seconds and then decelerates for \( t_2 \) seconds, we can relate \( t_1 \) and \( t_2 \) using the relationship of acceleration and retardation. The final velocity after acceleration will be:
\[ v = a \cdot t_1 \]
During deceleration, the initial velocity is \( v \) and the final velocity is 0. The equation for deceleration can be expressed as:
\[ 0 = v - d \cdot t_2 \]
Substituting \( d = 3a \) into the equation gives:
\[ 0 = a \cdot t_1 - 3a \cdot t_2 \]
Dividing through by \( a \) (assuming \( a \neq 0 \)):
\[ 0 = t_1 - 3t_2 \]
Thus, we can express \( t_1 \) in terms of \( t_2 \):
\[ t_1 = 3t_2 \]

### Step 4: Substitute into Total Time Equation
Now substitute \( t_1 = 3t_2 \) into the total time equation:
\[ 3t_2 + t_2 = 8 \]
This simplifies to:
\[ 4t_2 = 8 \]

### Step 5: Solve for \( t_2 \)
Now, solve for \( t_2 \):
\[ t_2 = \frac{8}{4} = 2 \text{ seconds} \]

### Step 6: Find \( t_1 \)
Now substitute \( t_2 = 2 \) back into the equation for \( t_1 \):
\[ t_1 = 3t_2 = 3 \cdot 2 = 6 \text{ seconds} \]

### Final Answer
The time for which the particle accelerates is:
\[ t_1 = 6 \text{ seconds} \]

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