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 the maximum velocity aquired by the car is

Options

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

Answer

To solve the problem, we will break it down into steps to find the maximum velocity acquired by the car during its motion.

### Step 1: Understand the Motion Phases
The car undergoes two phases:
1. **Acceleration Phase**: The car accelerates from rest at a constant rate `alpha` for a time `t1`.
2. **Deceleration Phase**: The car decelerates at a constant rate `beta` for a time `t2` until it comes to rest.

### Step 2: Define Variables
- Let the maximum velocity acquired by the car be `v`.
- The total time of motion is given as `t = t1 + t2`.

### Step 3: Write Equations for Each Phase
1. **For the Acceleration Phase**:
Using the formula for velocity under constant acceleration:
\[ v = 0 + \alpha t1 \quad \text{(initial velocity is 0)} \]
Thus, we can express `t1` in terms of `v`:
\[ t1 = \frac{v}{\alpha} \quad \text{(Equation 1)} \]

2. **For the Deceleration Phase**:
Using the formula for velocity under constant deceleration:
\[ 0 = v - \beta t2 \quad \text{(final velocity is 0)} \]
Rearranging gives us:
\[ t2 = \frac{v}{\beta} \quad \text{(Equation 2)} \]

### Step 4: Substitute into Total Time Equation
From the total time equation:
\[ t = t1 + t2 \]
Substituting the expressions for `t1` and `t2`:
\[ t = \frac{v}{\alpha} + \frac{v}{\beta} \]

### Step 5: Combine the Terms
Factoring out `v` from the right-hand side:
\[ t = v \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \]
To combine the fractions:
\[ t = v \left(\frac{\beta + \alpha}{\alpha \beta}\right) \]

### Step 6: Solve for Maximum Velocity `v`
Rearranging the equation to solve for `v`:
\[ v = t \cdot \frac{\alpha \beta}{\alpha + \beta} \]

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