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, the maximum velocity acquired by the car is given by :

Options

A. `((alpha^2-beta^2)/(alphabeta))t`
B. `((alpha^2+beta^2)/(alphabeta))t`
C. `((alpha+beta)t)/(alphabeta)`
D. `(alphabetat)/(alpha+beta)`

Answer

To solve the problem step by step, we will analyze the motion of the car during its acceleration and deceleration phases.

### Step 1: Define Variables
Let:
- \( \alpha \) = acceleration of the car (m/s²)
- \( \beta \) = deceleration of the car (m/s²)
- \( t_1 \) = time taken to accelerate to maximum velocity (s)
- \( t_2 \) = time taken to decelerate to rest (s)
- \( v \) = maximum velocity acquired by the car (m/s)
- \( t \) = total time elapsed (s)

### Step 2: Relationship during Acceleration
Since the car starts from rest, we can use the equation of motion:
\[ v = u + \alpha t_1 \]
Here, \( u = 0 \) (initial velocity), so:
\[ v = \alpha t_1 \tag{1} \]

### Step 3: Relationship during Deceleration
When the car decelerates to come to rest, we can use the same equation of motion:
\[ 0 = v - \beta t_2 \]
Rearranging gives:
\[ v = \beta t_2 \tag{2} \]

### Step 4: Total Time Relation
The total time \( t \) is the sum of the time taken to accelerate and the time taken to decelerate:
\[ t = t_1 + t_2 \tag{3} \]

### Step 5: Substitute for \( t_1 \) and \( t_2 \)
From equations (1) and (2), we can express \( t_1 \) and \( t_2 \) in terms of \( v \):
From (1):
\[ t_1 = \frac{v}{\alpha} \tag{4} \]
From (2):
\[ t_2 = \frac{v}{\beta} \tag{5} \]

### Step 6: Substitute \( t_1 \) and \( t_2 \) into Total Time Equation
Substituting equations (4) and (5) into equation (3):
\[ t = \frac{v}{\alpha} + \frac{v}{\beta} \]
Factoring out \( v \):
\[ t = v \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) \]

### Step 7: Solve for Maximum Velocity \( v \)
Rearranging the equation to solve for \( v \):
\[ v = \frac{t}{\left( \frac{1}{\alpha} + \frac{1}{\beta} \right)} \]
This can be simplified to:
\[ v = \frac{t \alpha \beta}{\alpha + \beta} \]

### Final Result
Thus, the maximum velocity acquired by the car is given by:
\[ v = \frac{\alpha \beta t}{\alpha + \beta} \]