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 (a) the maximum velocity reached and (b) the total distance travelled.

Options

A. `(alpha beta t^(2))/(4(alpha +beta))`
B. ` (alpha beta t^(2))/(2(alpha +beta))`
C. `(alpha beta t^(2))/(alpha +beta)`
D. `(4alpha beta t^(2))/(alpha +beta)`

Explanation

To solve the problem step by step, we will break it down into two parts: finding the maximum velocity reached and calculating the total distance traveled.

### Step 1: Define Variables
Let:
- \( \alpha \) = acceleration of the car
- \( \beta \) = deceleration of the car
- \( T \) = total time of motion
- \( T_1 \) = time during which the car accelerates
- \( T_2 \) = time during which the car decelerates
- \( V_{\text{max}} \) = maximum velocity reached

### Step 2: Relate Times
Since the total time is the sum of the time accelerating and the time decelerating, we have:
\[ T = T_1 + T_2 \]
Thus, we can express \( T_2 \) as:
\[ T_2 = T - T_1 \]

### Step 3: Calculate Maximum Velocity
During the acceleration phase, the car starts from rest, so the maximum velocity reached after time \( T_1 \) is given by:
\[ V_{\text{max}} = \alpha T_1 \]

During the deceleration phase, the car comes to rest from its maximum velocity \( V_{\text{max}} \) in time \( T_2 \). The equation for the deceleration phase is:
\[ V_{\text{max}} = \beta T_2 \]

Substituting \( T_2 \) from Step 2:
\[ V_{\text{max}} = \beta (T - T_1) \]

### Step 4: Set the Two Expressions for Maximum Velocity Equal
Since both expressions represent the maximum velocity, we can set them equal:
\[ \alpha T_1 = \beta (T - T_1) \]

### Step 5: Solve for \( T_1 \)
Rearranging the equation gives:
\[ \alpha T_1 + \beta T_1 = \beta T \]
\[ T_1 (\alpha + \beta) = \beta T \]
Thus,
\[ T_1 = \frac{\beta T}{\alpha + \beta} \]

### Step 6: Substitute \( T_1 \) Back to Find \( V_{\text{max}} \)
Now substituting \( T_1 \) back into the equation for maximum velocity:
\[ V_{\text{max}} = \alpha T_1 = \alpha \left(\frac{\beta T}{\alpha + \beta}\right) \]
This simplifies to:
\[ V_{\text{max}} = \frac{\alpha \beta T}{\alpha + \beta} \]

### Step 7: Calculate Distance Traveled During Acceleration
Using the equation of motion:
\[ S_1 = ut + \frac{1}{2} a t^2 \]
Here, initial velocity \( u = 0 \), acceleration \( a = \alpha \), and time \( t = T_1 \):
\[ S_1 = 0 + \frac{1}{2} \alpha T_1^2 \]
Substituting \( T_1 \):
\[ S_1 = \frac{1}{2} \alpha \left(\frac{\beta T}{\alpha + \beta}\right)^2 \]
This simplifies to:
\[ S_1 = \frac{\alpha \beta^2 T^2}{2(\alpha + \beta)^2} \]

### Step 8: Calculate Distance Traveled During Deceleration
Using the same equation of motion for the deceleration phase:
\[ S_2 = V_{\text{max}} T_2 - \frac{1}{2} \beta T_2^2 \]
Substituting \( V_{\text{max}} \) and \( T_2 \):
\[ S_2 = \frac{\alpha \beta T}{\alpha + \beta} (T - T_1) - \frac{1}{2} \beta (T - T_1)^2 \]
After substituting and simplifying, we find:
\[ S_2 = \frac{\beta \alpha^2 T^2}{2(\alpha + \beta)^2} \]

### Step 9: Total Distance Traveled
The total distance \( S \) is the sum of \( S_1 \) and \( S_2 \):
\[ S = S_1 + S_2 \]
\[ S = \frac{\alpha \beta^2 T^2}{2(\alpha + \beta)^2} + \frac{\beta \alpha^2 T^2}{2(\alpha + \beta)^2} \]
Factoring out common terms:
\[ S = \frac{\alpha \beta T^2 ( \beta + \alpha)}{2(\alpha + \beta)^2} \]
This simplifies to:
\[ S = \frac{\alpha \beta T^2}{2(\alpha + \beta)} \]

### Final Answers
(a) The maximum velocity reached is:
\[ V_{\text{max}} = \frac{\alpha \beta T}{\alpha + \beta} \]

(b) The total distance traveled is:
\[ S = \frac{\alpha \beta T^2}{2(\alpha + \beta)} \]