A car accelerates from rest at a constant rate `alpha` for some time, after which it decelerates at a constant rate `beta` and ultimately comes to rest.If the total time-lapse is t, what is the total distance described and maximum velocity reached ?

Answer

To solve the problem, we will break it down into steps to find the total distance traveled by the car and the maximum velocity reached.

### Step 1: Define Variables
Let:
- \( \alpha \) = acceleration of the car
- \( \beta \) = deceleration of the car
- \( t \) = total time of travel
- \( t_1 \) = time of acceleration
- \( t_2 \) = time of deceleration
- \( v_{max} \) = maximum velocity reached

### Step 2: Relate Time of Acceleration and Deceleration
Since the car accelerates for time \( t_1 \) and decelerates for time \( t_2 \), we have:
\[ t_1 + t_2 = t \]

### Step 3: Find Maximum Velocity
During acceleration, the maximum velocity \( v_{max} \) can be expressed as:
\[ v_{max} = \alpha t_1 \]

During deceleration, the relationship between initial velocity, final velocity, and time can be used:
\[ 0 = v_{max} - \beta t_2 \implies v_{max} = \beta t_2 \]

Equating the two expressions for \( v_{max} \):
\[ \alpha t_1 = \beta t_2 \]

From this, we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = \frac{\alpha}{\beta} t_1 \]

### Step 4: Substitute \( t_2 \) into Total Time Equation
Substituting \( t_2 \) into the total time equation:
\[ t_1 + \frac{\alpha}{\beta} t_1 = t \]
\[ t_1 \left(1 + \frac{\alpha}{\beta}\right) = t \]
\[ t_1 = \frac{\beta t}{\alpha + \beta} \]

### Step 5: Find \( t_2 \)
Now substitute \( t_1 \) back to find \( t_2 \):
\[ t_2 = \frac{\alpha}{\beta} t_1 = \frac{\alpha}{\beta} \cdot \frac{\beta t}{\alpha + \beta} = \frac{\alpha t}{\alpha + \beta} \]

### Step 6: Calculate Total Distance
The total distance \( D \) can be calculated as the sum of the distance covered during acceleration and deceleration.

1. Distance during acceleration:
\[ D_1 = \frac{1}{2} \alpha t_1^2 = \frac{1}{2} \alpha \left(\frac{\beta t}{\alpha + \beta}\right)^2 = \frac{\alpha \beta^2 t^2}{2(\alpha + \beta)^2} \]

2. Distance during deceleration:
\[ D_2 = v_{max} t_2 - \frac{1}{2} \beta t_2^2 \]
Substituting \( v_{max} \) and \( t_2 \):
\[ D_2 = \alpha t_1 \cdot \frac{\alpha t}{\alpha + \beta} - \frac{1}{2} \beta \left(\frac{\alpha t}{\alpha + \beta}\right)^2 \]
\[ D_2 = \frac{\alpha^2 \beta t}{\alpha + \beta} - \frac{1}{2} \beta \cdot \frac{\alpha^2 t^2}{(\alpha + \beta)^2} \]

### Step 7: Combine Distances
Now, combine \( D_1 \) and \( D_2 \):
\[ D = D_1 + D_2 = \frac{\alpha \beta^2 t^2}{2(\alpha + \beta)^2} + \left(\frac{\alpha^2 \beta t}{\alpha + \beta} - \frac{1}{2} \beta \cdot \frac{\alpha^2 t^2}{(\alpha + \beta)^2}\right) \]

After simplification:
\[ D = \frac{\alpha \beta t^2}{2(\alpha + \beta)} \]

### Step 8: Find Maximum Velocity
Substituting \( t_1 \) back into the expression for maximum velocity:
\[ v_{max} = \alpha t_1 = \alpha \cdot \frac{\beta t}{\alpha + \beta} = \frac{\alpha \beta t}{\alpha + \beta} \]

### Final Results
- Total Distance \( D = \frac{\alpha \beta t^2}{2(\alpha + \beta)} \)
- Maximum Velocity \( v_{max} = \frac{\alpha \beta t}{\alpha + \beta} \)