A car accelerates from rest at a constant rate `alpha` for some time, after which it decelerates at a constant rate `beta,` to come to rest. If the total time elapsed is t seconds. Then evalute (a) the maximum velocity reached and (b) the total distance travelled.

Explanation

To solve the problem step by step, we will evaluate both the maximum velocity reached by the car and the total distance traveled during the acceleration and deceleration phases.

### Step 1: Understanding the Motion
The car accelerates from rest at a constant rate \( \alpha \) for a time \( t_1 \) and then decelerates at a constant rate \( \beta \) for a time \( t_2 \) until it comes to rest. The total time of the journey is given as \( t = t_1 + t_2 \).

### Step 2: Finding Maximum Velocity
1. **From A to B (Acceleration Phase)**:
- Initial velocity \( u = 0 \)
- Final velocity \( v = v_{max} \)
- Acceleration \( a = \alpha \)
- Time \( t_1 \)

Using the first equation of motion:
\[ v_{max} = u + \alpha t_1 \implies v_{max} = 0 + \alpha t_1 \implies v_{max} = \alpha t_1 \]

2. **From B to C (Deceleration Phase)**:
- Initial velocity \( u = v_{max} \)
- Final velocity \( v = 0 \)
- Deceleration \( a = -\beta \)
- Time \( t_2 \)

Using the first equation of motion:
\[ 0 = v_{max} - \beta t_2 \implies v_{max} = \beta t_2 \]

3. **Equating the Two Expressions for \( v_{max} \)**:
From the two equations we derived:
\[ \alpha t_1 = \beta t_2 \]

4. **Using Total Time**:
Since \( t = t_1 + t_2 \), we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = t - t_1 \]

5. **Substituting for \( t_2 \)**:
Substitute \( t_2 \) into the equation \( \alpha t_1 = \beta t_2 \):
\[ \alpha t_1 = \beta (t - t_1) \]
Rearranging gives:
\[ \alpha t_1 + \beta t_1 = \beta t \implies t_1 (\alpha + \beta) = \beta t \implies t_1 = \frac{\beta t}{\alpha + \beta} \]

6. **Finding \( t_2 \)**:
Substitute \( t_1 \) back to find \( t_2 \):
\[ t_2 = t - t_1 = t - \frac{\beta t}{\alpha + \beta} = \frac{\alpha t}{\alpha + \beta} \]

7. **Substituting Back to Find \( v_{max} \)**:
Now substitute \( t_1 \) into \( v_{max} \):
\[ v_{max} = \alpha t_1 = \alpha \left(\frac{\beta t}{\alpha + \beta}\right) = \frac{\alpha \beta t}{\alpha + \beta} \]

### Step 3: Finding Total Distance
1. **Distance from A to B**:
Using the equation of motion:
\[ s_1 = ut + \frac{1}{2} a t_1^2 = 0 + \frac{1}{2} \alpha t_1^2 = \frac{1}{2} \alpha \left(\frac{\beta t}{\alpha + \beta}\right)^2 \]
Simplifying gives:
\[ s_1 = \frac{1}{2} \alpha \frac{\beta^2 t^2}{(\alpha + \beta)^2} \]

2. **Distance from B to C**:
Using the equation of motion:
\[ s_2 = v_{max} t_2 - \frac{1}{2} \beta t_2^2 \]
Substituting \( v_{max} \) and \( t_2 \):
\[ s_2 = \frac{\alpha \beta t}{\alpha + \beta} \cdot \frac{\alpha t}{\alpha + \beta} - \frac{1}{2} \beta \left(\frac{\alpha t}{\alpha + \beta}\right)^2 \]
Simplifying gives:
\[ s_2 = \frac{\alpha \beta t^2}{(\alpha + \beta)^2} - \frac{1}{2} \beta \frac{\alpha^2 t^2}{(\alpha + \beta)^2} = \frac{(\alpha \beta - \frac{1}{2} \alpha^2) t^2}{(\alpha + \beta)^2} \]

3. **Total Distance**:
The total distance \( s \) is:
\[ s = s_1 + s_2 = \frac{1}{2} \frac{\alpha \beta^2 t^2}{(\alpha + \beta)^2} + \frac{(\alpha \beta - \frac{1}{2} \alpha^2) t^2}{(\alpha + \beta)^2} \]
Combining gives:
\[ s = \frac{t^2}{2(\alpha + \beta)} \left(\frac{\alpha \beta + \alpha \beta - \frac{1}{2} \alpha^2}{(\alpha + \beta)}\right) = \frac{t^2}{2(\alpha + \beta)} \left(\frac{2\alpha \beta - \frac{1}{2} \alpha^2}{(\alpha + \beta)}\right) \]

### Final Answers
(a) The maximum velocity reached is:
\[ v_{max} = \frac{\alpha \beta t}{\alpha + \beta} \]

(b) The total distance traveled is:
\[ s = \frac{t^2}{2(\alpha + \beta)} \]