The engine of a motoecycle can produce a maximum acceleration of `5 m//s^(2)`. Its brakes can produce a maximum retardation of `10 m//s^(2)`. What is the minimum time in which the motorcycle can cover a distance of `1.5 km`?

Options

A. (1)`30s`
B. (2)`15 s`
C. (3)`10 s`
D. (4)`5 s`

Explanation

To solve the problem, we need to find the minimum time in which the motorcycle can cover a distance of 1.5 km (or 1500 m) given the maximum acceleration and retardation.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Maximum acceleration, \( a = 5 \, \text{m/s}^2 \)
- Maximum retardation, \( b = 10 \, \text{m/s}^2 \)
- Distance, \( s = 1500 \, \text{m} \)

2. **Understanding the Motion:**
- The motorcycle will first accelerate to a certain maximum speed and then will have to decelerate to stop. To minimize the time, we need to maximize the speed reached during the acceleration phase.

3. **Using the equations of motion:**
- The distance covered during acceleration can be expressed as:
\[ s_1 = \frac{1}{2} a t_1^2 \]
where \( t_1 \) is the time taken to accelerate.

- The maximum speed reached at the end of the acceleration phase is:
\[ v = a t_1 \]

4. **Distance during Deceleration:**
- The distance covered during deceleration can be expressed as:
\[ s_2 = \frac{v^2}{2b} \]
- Substituting \( v = a t_1 \) into the equation for \( s_2 \):
\[ s_2 = \frac{(a t_1)^2}{2b} = \frac{a^2 t_1^2}{2b} \]

5. **Total Distance:**
- The total distance covered is:
\[ s = s_1 + s_2 \]
- Substituting the expressions for \( s_1 \) and \( s_2 \):
\[ s = \frac{1}{2} a t_1^2 + \frac{a^2 t_1^2}{2b} \]

6. **Setting up the equation:**
- Plugging in the values of \( a \) and \( b \):
\[ 1500 = \frac{1}{2} (5) t_1^2 + \frac{(5)^2 t_1^2}{2(10)} \]
- Simplifying:
\[ 1500 = \frac{5}{2} t_1^2 + \frac{25}{20} t_1^2 = \frac{5}{2} t_1^2 + \frac{5}{4} t_1^2 \]
- Finding a common denominator (4):
\[ 1500 = \frac{10}{4} t_1^2 + \frac{5}{4} t_1^2 = \frac{15}{4} t_1^2 \]

7. **Solving for \( t_1^2 \):**
- Rearranging gives:
\[ t_1^2 = \frac{1500 \times 4}{15} = 400 \]
- Thus, \( t_1 = \sqrt{400} = 20 \, \text{s} \)

8. **Total Time:**
- The total time \( t \) is the time to accelerate \( t_1 \) plus the time to decelerate \( t_2 \):
- The time to decelerate can be found using:
\[ t_2 = \frac{v}{b} = \frac{a t_1}{b} = \frac{5 \times 20}{10} = 10 \, \text{s} \]
- Therefore, the total time \( t = t_1 + t_2 = 20 + 10 = 30 \, \text{s} \)

### Final Answer:
The minimum time in which the motorcycle can cover a distance of 1.5 km is **30 seconds**.