A car accelerates from rest at a constant rate `alpha` for sometime , after which it decelerates at a constant rate `beta` and comes to rest. If T is the total time elapsed, the maximum velocity acquired by the car is

Options

A. `((alphabeta)/(alpha+beta))T`
B. `((alpha+beta/(alphabeta))T`
C. `((alpha^(2)+beta^(2))/(alphabeta))T`
D. `(1)/(2)(alphaT)/(beta)`

Explanation

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

### Step 1: Understand the Motion Phases
The car accelerates from rest at a constant rate `α` for a time `t1`, reaches a maximum velocity `Vmax`, and then decelerates at a constant rate `β` for a time `t2` until it comes to rest.

### Step 2: Apply the First Equation of Motion for Acceleration
Using the first equation of motion:
\[ V = U + at \]
Where:
- \( V \) is the final velocity (which is `Vmax`),
- \( U \) is the initial velocity (which is `0` since the car starts from rest),
- \( a \) is the acceleration (`α`),
- \( t \) is the time of acceleration (`t1`).

Substituting the known values:
\[ Vmax = 0 + α \cdot t1 \]
This simplifies to:
\[ Vmax = α \cdot t1 \]
From this, we can express `t1` as:
\[ t1 = \frac{Vmax}{α} \]

### Step 3: Apply the First Equation of Motion for Deceleration
Now, during the deceleration phase, we again use the first equation of motion:
\[ V = U + at \]
Where:
- \( V \) is the final velocity (which is `0` since the car comes to rest),
- \( U \) is the initial velocity (which is `Vmax`),
- \( a \) is the deceleration (which is `-β`),
- \( t \) is the time of deceleration (`t2`).

Substituting the known values:
\[ 0 = Vmax - β \cdot t2 \]
Rearranging gives:
\[ β \cdot t2 = Vmax \]
From this, we can express `t2` as:
\[ t2 = \frac{Vmax}{β} \]

### Step 4: Relate Total Time to Acceleration and Deceleration Times
The total time `T` is the sum of the acceleration time `t1` and the deceleration time `t2`:
\[ T = t1 + t2 \]
Substituting the expressions for `t1` and `t2`:
\[ T = \frac{Vmax}{α} + \frac{Vmax}{β} \]

### Step 5: Factor Out `Vmax`
Factoring out `Vmax` from the equation:
\[ T = Vmax \left( \frac{1}{α} + \frac{1}{β} \right) \]
This can be rewritten as:
\[ T = Vmax \cdot \frac{α + β}{αβ} \]

### Step 6: Solve for `Vmax`
Now, we can solve for `Vmax`:
\[ Vmax = \frac{T \cdot αβ}{α + β} \]

### Final Expression for Maximum Velocity
Thus, the maximum velocity acquired by the car is:
\[ Vmax = \frac{αβ}{α + β} \cdot T \]