A car accelerates from rest at a constant rate 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 beta )/( alpha + beta ) t`
B. ` ( alpha + beta )/( alpha beta ) t`
C. ` ( alpha ^(2) + beta ^(2))/( beta alpha ) `
D. ` (alpha + beta^(2))/( beta alpha ) `

Explanation

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

### Step 1: Define the Variables
- Let \( \alpha \) be the constant acceleration of the car.
- Let \( \beta \) be the constant deceleration of the car.
- Let \( t_1 \) be the time during which the car accelerates.
- Let \( t_2 \) be the time during which the car decelerates.
- The total time elapsed is \( t = t_1 + t_2 \).

### Step 2: Determine the Maximum Velocity
During the acceleration phase, the car starts from rest and accelerates to a maximum velocity \( v_{\text{max}} \). The relationship between acceleration, time, and velocity is given by:
\[ v_{\text{max}} = \alpha t_1 \]

### Step 3: Determine the Deceleration Phase
During the deceleration phase, the car comes to rest from the maximum velocity \( v_{\text{max}} \). The relationship for deceleration is:
\[ v_{\text{max}} = \beta t_2 \]
Since the car comes to rest, we can express \( t_2 \) in terms of \( v_{\text{max}} \):
\[ t_2 = \frac{v_{\text{max}}}{\beta} \]

### Step 4: Relate the Times
From the total time equation, we have:
\[ t = t_1 + t_2 \]
Substituting \( t_2 \) from the deceleration phase:
\[ t = t_1 + \frac{v_{\text{max}}}{\beta} \]

### Step 5: Substitute \( v_{\text{max}} \)
Now, substitute \( v_{\text{max}} = \alpha t_1 \) into the equation:
\[ t = t_1 + \frac{\alpha t_1}{\beta} \]
Factoring \( t_1 \) out:
\[ t = t_1 \left(1 + \frac{\alpha}{\beta}\right) \]

### Step 6: Solve for \( t_1 \)
Rearranging gives:
\[ t_1 = \frac{t}{1 + \frac{\alpha}{\beta}} = \frac{t \beta}{\beta + \alpha} \]

### Step 7: Substitute Back to Find \( v_{\text{max}} \)
Now substitute \( t_1 \) back into the equation for \( v_{\text{max}} \):
\[ v_{\text{max}} = \alpha t_1 = \alpha \left(\frac{t \beta}{\beta + \alpha}\right) \]
Thus, we have:
\[ v_{\text{max}} = \frac{\alpha \beta t}{\beta + \alpha} \]

### Final Answer
The maximum velocity acquired by the car is given by:
\[ v_{\text{max}} = \frac{\alpha \beta t}{\alpha + \beta} \]