A car is travelling on a straight road. The maximum velocity the car Can attain is `24 ms^-1.` The maximum acceleration and deceleration it can attain are `1 ms^-2` and `4 ms^-2` respectively. The shortest time the car takes from rest to rest in a distance of 200 m is,

Options

A. `22.4 s` (b) 30 s (c) 11.2 s (d) 5.6 s
B. `30 s`
C. `11.2 s`
D. `5.6 s`

Explanation

To solve the problem step by step, we will analyze the motion of the car as it accelerates and decelerates.

### Step 1: Understand the motion phases
The car starts from rest, accelerates to its maximum velocity, then decelerates back to rest. We denote:
- Maximum velocity \( V_{max} = 24 \, \text{m/s} \)
- Maximum acceleration \( a = 1 \, \text{m/s}^2 \)
- Maximum deceleration \( b = 4 \, \text{m/s}^2 \)

### Step 2: Determine the time taken to reach maximum velocity
Using the formula for acceleration:
\[ V = u + at \]
where \( u = 0 \) (initial velocity), we can find the time \( t_a \) taken to reach maximum velocity:
\[ 24 = 0 + 1 \cdot t_a \implies t_a = 24 \, \text{s} \]

### Step 3: Calculate the distance covered during acceleration
The distance \( s_a \) covered during the acceleration phase can be calculated using:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting \( u = 0 \):
\[ s_a = 0 + \frac{1}{2} \cdot 1 \cdot t_a^2 = \frac{1}{2} \cdot 1 \cdot (24)^2 = 288 \, \text{m} \]

### Step 4: Determine the time taken to decelerate
Since the car decelerates at \( b = 4 \, \text{m/s}^2 \), the time \( t_d \) taken to come to rest from \( V_{max} \) is:
\[ 0 = 24 - 4 t_d \implies t_d = \frac{24}{4} = 6 \, \text{s} \]

### Step 5: Calculate the distance covered during deceleration
Using the same formula for distance:
\[ s_d = V_{max} t_d - \frac{1}{2} b t_d^2 \]
Substituting the values:
\[ s_d = 24 \cdot 6 - \frac{1}{2} \cdot 4 \cdot (6)^2 = 144 - 72 = 72 \, \text{m} \]

### Step 6: Total distance and time
The total distance covered during both phases is:
\[ s_{total} = s_a + s_d = 288 + 72 = 360 \, \text{m} \]
However, we need to cover only 200 m. Thus, we need to adjust the time spent accelerating and decelerating.

### Step 7: Set up the equations for the total distance
Let \( t \) be the time spent accelerating. The distance covered during acceleration is:
\[ s_a = \frac{1}{2} \cdot 1 \cdot t^2 = \frac{t^2}{2} \]
The time spent decelerating is \( \frac{t}{4} \) (as given in the problem). The distance during deceleration is:
\[ s_d = 24 \cdot \frac{t}{4} - \frac{1}{2} \cdot 4 \cdot \left(\frac{t}{4}\right)^2 = 6t - \frac{1}{2} \cdot 4 \cdot \frac{t^2}{16} = 6t - \frac{t^2}{8} \]

### Step 8: Total distance equation
Setting the total distance to 200 m:
\[ \frac{t^2}{2} + 6t - \frac{t^2}{8} = 200 \]
Multiplying through by 8 to eliminate the fraction:
\[ 4t^2 + 48t - t^2 = 1600 \]
This simplifies to:
\[ 3t^2 + 48t - 1600 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-48 \pm \sqrt{48^2 - 4 \cdot 3 \cdot (-1600)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ t = \frac{-48 \pm \sqrt{2304 + 19200}}{6} \]
\[ t = \frac{-48 \pm \sqrt{21504}}{6} \]
Calculating \( \sqrt{21504} \approx 146.6 \):
\[ t = \frac{-48 + 146.6}{6} \approx 16.433 \, \text{s} \]

### Step 10: Total time for the journey
The total time taken is:
\[ T = t + \frac{t}{4} = t \left(1 + \frac{1}{4}\right) = \frac{5t}{4} \approx \frac{5 \cdot 16.433}{4} \approx 20.54 \, \text{s} \]

### Final Answer
The shortest time the car takes from rest to rest in a distance of 200 m is approximately **20.54 seconds**.