The maximum acceleration that can be produced by the engine of a vehicle is `5ms^(-2)` and the maximum retardation is brake can produce is `10ms^(-2)`. Then the minimum time during which it can cover a distance of 60 m is

Options

A. 6 s
B. 5 s
C. 36 s
D. 24 s.

Explanation

To solve the problem, we need to determine the minimum time required for a vehicle to cover a distance of 60 meters given its maximum acceleration and maximum retardation.

### Step-by-Step Solution:

1. **Define the Variables:**
- Maximum acceleration, \( a_1 = 5 \, \text{m/s}^2 \)
- Maximum retardation, \( a_2 = -10 \, \text{m/s}^2 \) (negative because it is deceleration)
- Total distance, \( S = 60 \, \text{m} \)

2. **Distance During Acceleration:**
- Let \( s_1 \) be the distance covered during the acceleration phase.
- Using the kinematic equation:
\[ V^2 = U^2 + 2a_1 s_1 \]
where \( U = 0 \) (initial velocity during acceleration), we get:
\[ V^2 = 2 \cdot 5 \cdot s_1 \implies V^2 = 10 s_1 \]

3. **Distance During Retardation:**
- Let \( s_2 \) be the distance covered during the retardation phase.
- Using the kinematic equation:
\[ V^2 = U^2 + 2a_2 s_2 \]
where \( U = V \) (initial velocity during retardation) and \( V = 0 \) at the end of retardation, we have:
\[ 0 = V^2 - 2 \cdot 10 \cdot s_2 \implies V^2 = 20 s_2 \]

4. **Total Distance:**
- The total distance covered is the sum of the distances during acceleration and retardation:
\[ S = s_1 + s_2 = 60 \]
- Substitute \( s_1 \) and \( s_2 \) in terms of \( V \):
\[ s_1 = \frac{V^2}{10}, \quad s_2 = \frac{V^2}{20} \]
- Therefore, we can write:
\[ \frac{V^2}{10} + \frac{V^2}{20} = 60 \]

5. **Combine the Terms:**
- Find a common denominator (which is 20):
\[ \frac{2V^2}{20} + \frac{V^2}{20} = 60 \implies \frac{3V^2}{20} = 60 \]

6. **Solve for \( V^2 \):**
- Multiply both sides by 20:
\[ 3V^2 = 1200 \implies V^2 = 400 \implies V = 20 \, \text{m/s} \]

7. **Calculate Time During Acceleration:**
- Using the equation \( V = U + a_1 t_1 \):
\[ 20 = 0 + 5 t_1 \implies t_1 = \frac{20}{5} = 4 \, \text{s} \]

8. **Calculate Time During Retardation:**
- Using the equation \( V = U + a_2 t_2 \):
\[ 0 = 20 - 10 t_2 \implies 10 t_2 = 20 \implies t_2 = 2 \, \text{s} \]

9. **Total Time:**
- The total time \( T \) to cover the distance of 60 m is:
\[ T = t_1 + t_2 = 4 + 2 = 6 \, \text{s} \]

### Final Answer:
The minimum time during which the vehicle can cover a distance of 60 m is **6 seconds**.