A car moves in a straight line, the car accelerates from rest with a constant acceleraation `alpha` on a straight foad. After gaining a velocity `v`, the car moves with that velocity for somerime. Then car decelerates with a retardation `beta`, If the total distance covered by the car is equal to `s` find the total time of its motion.

Explanation

To solve the problem, we need to break down the motion of the car into three distinct phases: acceleration, constant velocity, and deceleration. We will then use the equations of motion and the relationships between distance, velocity, and time to find the total time of motion.

### Step-by-Step Solution:

1. **Phase 1: Acceleration**
- The car accelerates from rest with a constant acceleration \( \alpha \).
- Initial velocity \( u = 0 \).
- Final velocity \( v \) is reached after time \( t_1 \).
- Using the equation of motion:
\[ v = u + \alpha t_1 \implies v = 0 + \alpha t_1 \implies t_1 = \frac{v}{\alpha} \]
- The distance covered during this phase \( s_1 \) can be calculated using:
\[ s_1 = ut + \frac{1}{2} \alpha t_1^2 = 0 + \frac{1}{2} \alpha t_1^2 = \frac{1}{2} \alpha \left(\frac{v}{\alpha}\right)^2 = \frac{v^2}{2\alpha} \]

2. **Phase 2: Constant Velocity**
- The car moves with constant velocity \( v \) for a time \( t_2 \).
- The distance covered during this phase \( s_2 \) is:
\[ s_2 = v t_2 \]

3. **Phase 3: Deceleration**
- The car decelerates with a retardation \( \beta \) until it comes to rest.
- The time taken to decelerate \( t_3 \) can be found using:
\[ v = 0 + \beta t_3 \implies t_3 = \frac{v}{\beta} \]
- The distance covered during this phase \( s_3 \) is:
\[ s_3 = vt_3 - \frac{1}{2} \beta t_3^2 = v\left(\frac{v}{\beta}\right) - \frac{1}{2} \beta \left(\frac{v}{\beta}\right)^2 = \frac{v^2}{\beta} - \frac{1}{2} \frac{v^2}{\beta} = \frac{v^2}{2\beta} \]

4. **Total Distance**
- The total distance \( s \) covered by the car is the sum of the distances from all three phases:
\[ s = s_1 + s_2 + s_3 = \frac{v^2}{2\alpha} + vt_2 + \frac{v^2}{2\beta} \]

5. **Rearranging for \( t_2 \)**
- Rearranging the equation for \( t_2 \):
\[ vt_2 = s - \left(\frac{v^2}{2\alpha} + \frac{v^2}{2\beta}\right) \]
\[ t_2 = \frac{s - \left(\frac{v^2}{2\alpha} + \frac{v^2}{2\beta}\right)}{v} \]

6. **Total Time of Motion**
- The total time \( T \) is the sum of the times for each phase:
\[ T = t_1 + t_2 + t_3 = \frac{v}{\alpha} + \frac{s - \left(\frac{v^2}{2\alpha} + \frac{v^2}{2\beta}\right)}{v} + \frac{v}{\beta} \]
- Simplifying this gives:
\[ T = \frac{s}{v} + \frac{v}{2}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \]

### Final Result:
The total time of motion \( T \) is given by:
\[ T = \frac{s}{v} + \frac{v}{2}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \]