A car acceleration from rest at a constant rate `2m//s^(2)` for some time. Then, it retards at a constant rate of `4 m//s^(2)` and comes to rest. If it remains motion for 3 second, then the maximum speed attained by the car is:-

Options

A. `2ms^(-1)`
B. `3ms^(-1)`
C. `4ms^(-1)`
D. `6ms^(-1)`

Answer

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

### Step 1: Define Variables
Let:
- \( v \) = maximum speed attained by the car (in m/s)
- \( t_1 \) = time of acceleration (in seconds)
- \( t_2 \) = time of retardation (in seconds)

### Step 2: Use the equations of motion
1. **Acceleration Phase**:
- The car starts from rest, so the initial velocity \( u = 0 \).
- The acceleration \( a = 2 \, \text{m/s}^2 \).
- Using the equation of motion:
\[ v = u + at \implies v = 0 + 2t_1 \implies v = 2t_1 \]
- Therefore, we can express \( t_1 \) in terms of \( v \):
\[ t_1 = \frac{v}{2} \]

2. **Retardation Phase**:
- The car comes to rest, so the final velocity \( v_f = 0 \).
- The initial velocity for this phase is \( v \) and the retardation \( a = -4 \, \text{m/s}^2 \).
- Using the equation of motion:
\[ v_f = v + (-4)t_2 \implies 0 = v - 4t_2 \implies 4t_2 = v \]
- Therefore, we can express \( t_2 \) in terms of \( v \):
\[ t_2 = \frac{v}{4} \]

### Step 3: Total Time of Motion
According to the problem, the total time for which the car remains in motion is 3 seconds:
\[ t_1 + t_2 = 3 \]
Substituting the expressions for \( t_1 \) and \( t_2 \):
\[ \frac{v}{2} + \frac{v}{4} = 3 \]

### Step 4: Solve for \( v \)
To solve this equation, we need a common denominator:
\[ \frac{2v}{4} + \frac{v}{4} = 3 \implies \frac{3v}{4} = 3 \]
Multiplying both sides by 4:
\[ 3v = 12 \implies v = \frac{12}{3} = 4 \, \text{m/s} \]

### Final Answer
The maximum speed attained by the car is \( \boxed{4 \, \text{m/s}} \).

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