A car accelerates from rest at constant rate of `2 ms^(-2)` for some time. Then it retards at a constant rate of `4 ms^(-2)` and comes to rest. It remains in motion for 3 seconds

Options

A. It is accelerated for 2 sec
B. Maximum velocity attained is `4 ms^(-1)`
C. Total distance covered is 6 m
D. Total distance covered is 10 m

Explanation

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

### Step 1: Determine the time of acceleration (t1)
The car accelerates from rest with a constant acceleration of \( a_1 = 2 \, \text{m/s}^2 \). We denote the time of acceleration as \( t_1 \).

Using the formula for velocity:
\[ v = u + a t \]
where:
- \( u = 0 \) (initial velocity, since it starts from rest)
- \( a = 2 \, \text{m/s}^2 \)
- \( t = t_1 \)

The velocity at the end of the acceleration phase is:
\[ v = 0 + 2 t_1 = 2 t_1 \]

### Step 2: Determine the time of deceleration (t2)
After accelerating, the car begins to decelerate at a constant rate of \( a_2 = -4 \, \text{m/s}^2 \). We denote the time of deceleration as \( t_2 \).

Using the formula for velocity during deceleration:
\[ v' = v + a t \]
where:
- \( v = 2 t_1 \) (the velocity at the end of the acceleration phase)
- \( a = -4 \, \text{m/s}^2 \)
- \( t = t_2 \)

At the end of the deceleration, the car comes to rest, so:
\[ 0 = 2 t_1 - 4 t_2 \]
Rearranging gives:
\[ 2 t_1 = 4 t_2 \quad \Rightarrow \quad t_1 = 2 t_2 \]

### Step 3: Total time of motion
The total time of motion is given as 3 seconds:
\[ t_1 + t_2 = 3 \]
Substituting \( t_1 = 2 t_2 \) into this equation:
\[ 2 t_2 + t_2 = 3 \quad \Rightarrow \quad 3 t_2 = 3 \quad \Rightarrow \quad t_2 = 1 \, \text{s} \]
Now substituting back to find \( t_1 \):
\[ t_1 = 2 t_2 = 2 \times 1 = 2 \, \text{s} \]

### Step 4: Calculate maximum velocity
The maximum velocity attained at the end of the acceleration phase is:
\[ v = 2 t_1 = 2 \times 2 = 4 \, \text{m/s} \]

### Step 5: Calculate distance covered during acceleration (h1)
Using the formula for distance:
\[ h_1 = ut + \frac{1}{2} a t^2 \]
where:
- \( u = 0 \)
- \( a = 2 \, \text{m/s}^2 \)
- \( t = t_1 = 2 \, \text{s} \)

Thus,
\[ h_1 = 0 + \frac{1}{2} \times 2 \times (2^2) = \frac{1}{2} \times 2 \times 4 = 4 \, \text{m} \]

### Step 6: Calculate distance covered during deceleration (h2)
Using the formula for distance again:
\[ h_2 = vt + \frac{1}{2} a t^2 \]
where:
- \( v = 4 \, \text{m/s} \) (initial velocity during deceleration)
- \( a = -4 \, \text{m/s}^2 \)
- \( t = t_2 = 1 \, \text{s} \)

Thus,
\[ h_2 = 4 \times 1 + \frac{1}{2} \times (-4) \times (1^2) = 4 - 2 = 2 \, \text{m} \]

### Step 7: Total distance covered
The total distance covered by the car is:
\[ h_{\text{total}} = h_1 + h_2 = 4 + 2 = 6 \, \text{m} \]

### Final Summary
- Time of acceleration \( t_1 = 2 \, \text{s} \)
- Time of deceleration \( t_2 = 1 \, \text{s} \)
- Maximum velocity \( v = 4 \, \text{m/s} \)
- Total distance covered \( h_{\text{total}} = 6 \, \text{m} \)