A car accelerates from rest at constant rate of `2ms^(-2)` for some time. Immediately after this, it retards at a constant rate of `4ms^(-2)` and comes to rest. The total time for which it remains in motion is 3 s. Taking the moment of start of motion as t = 0, answer the following questions.
What is the average speed of the car for the journey .
a. 2 m/s
b. 3 m/s
c. 1 m/s
d. Zero

Answer

To solve the problem step by step, we need to analyze the motion of the car in two phases: acceleration and deceleration.

### Step 1: Understanding the Motion Phases
The car accelerates from rest at a constant rate of \(2 \, \text{m/s}^2\) for some time \(t\), and then it decelerates at a constant rate of \(4 \, \text{m/s}^2\) until it comes to rest. The total time of motion is given as \(3 \, \text{s}\).

### Step 2: Define the Time Intervals
Let:
- \(t_1\) be the time of acceleration.
- \(t_2\) be the time of deceleration.

From the problem, we know:
\[ t_1 + t_2 = 3 \, \text{s} \]

### Step 3: Calculate the Final Velocity after Acceleration
Using the formula for final velocity under constant acceleration:
\[ v = u + at \]
Here, \(u = 0\) (initial velocity), \(a = 2 \, \text{m/s}^2\), and \(t = t_1\):
\[ v = 0 + 2t_1 = 2t_1 \]

### Step 4: Set Up the Equation for Deceleration
During deceleration, the car comes to rest, so the final velocity \(v = 0\). Using the formula:
\[ v = u + at \]
where \(u = 2t_1\) (the velocity at the end of acceleration), \(a = -4 \, \text{m/s}^2\), and \(t = t_2\):
\[ 0 = 2t_1 - 4t_2 \]
Rearranging gives:
\[ 2t_1 = 4t_2 \quad \Rightarrow \quad t_1 = 2t_2 \]

### Step 5: Substitute \(t_2\) in the Time Equation
Substituting \(t_1 = 2t_2\) into the total time equation:
\[ 2t_2 + t_2 = 3 \]
\[ 3t_2 = 3 \quad \Rightarrow \quad t_2 = 1 \, \text{s} \]
Now substituting back to find \(t_1\):
\[ t_1 = 2t_2 = 2 \cdot 1 = 2 \, \text{s} \]

### Step 6: Calculate Distances
1. **Distance during acceleration (\(s_1\))**:
Using the formula:
\[ s = ut + \frac{1}{2}at^2 \]
Here, \(u = 0\), \(a = 2 \, \text{m/s}^2\), and \(t = t_1 = 2 \, \text{s}\):
\[ s_1 = 0 + \frac{1}{2} \cdot 2 \cdot (2^2) = \frac{1}{2} \cdot 2 \cdot 4 = 4 \, \text{m} \]

2. **Distance during deceleration (\(s_2\))**:
Using the same formula, where \(u = 2t_1 = 4 \, \text{m/s}\) (final velocity from acceleration), \(a = -4 \, \text{m/s}^2\), and \(t = t_2 = 1 \, \text{s}\):
\[ s_2 = 4 \cdot 1 + \frac{1}{2} \cdot (-4) \cdot (1^2) = 4 - 2 = 2 \, \text{m} \]

### Step 7: Total Distance and Average Speed
Total distance \(s = s_1 + s_2 = 4 + 2 = 6 \, \text{m}\).

Average speed \(V_{avg}\) is given by:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{6 \, \text{m}}{3 \, \text{s}} = 2 \, \text{m/s} \]

### Final Answer
The average speed of the car for the journey is \(2 \, \text{m/s}\).