A particle accelerates from rest with uniform acceleration `alpha` then decelerates to rest witha constant deceleration `beta`. Find total displacement. Given time is T

Options

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

Answer

To solve the problem of finding the total displacement of a particle that accelerates from rest with uniform acceleration \( \alpha \) and then decelerates to rest with a constant deceleration \( \beta \) over a total time \( T \), we can break the problem down into a few clear steps.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle starts from rest, accelerates for a time \( t_1 \) with acceleration \( \alpha \), and then decelerates for a time \( t_2 \) with deceleration \( \beta \) until it comes to rest.
- The total time of motion is given as \( T \), so we have:
\[ t_1 + t_2 = T \]

2. **Finding the Maximum Velocity**:
- During the acceleration phase, the final velocity \( v \) at time \( t_1 \) can be expressed using the equation of motion:
\[ v = \alpha t_1 \]

3. **Finding Time During Deceleration**:
- During the deceleration phase, the time \( t_2 \) can be expressed in terms of the maximum velocity \( v \):
\[ v = \beta t_2 \]
- Rearranging gives:
\[ t_2 = \frac{v}{\beta} \]

4. **Substituting for Total Time**:
- Substitute \( t_1 \) and \( t_2 \) into the total time equation:
\[ t_1 + \frac{v}{\beta} = T \]
- From \( v = \alpha t_1 \), we can express \( t_1 \) as:
\[ t_1 = \frac{v}{\alpha} \]
- Substitute this into the total time equation:
\[ \frac{v}{\alpha} + \frac{v}{\beta} = T \]

5. **Finding Maximum Velocity**:
- Factor out \( v \):
\[ v \left( \frac{1}{\alpha} + \frac{1}{\beta} \right) = T \]
- Solving for \( v \) gives:
\[ v = \frac{\alpha \beta T}{\alpha + \beta} \]

6. **Calculating Total Displacement**:
- The total displacement \( s \) can be calculated as the sum of the areas of the two triangles formed in the velocity-time graph:
\[ s = \text{Area of acceleration phase} + \text{Area of deceleration phase} \]
- The area of the first triangle (acceleration phase) is:
\[ s_1 = \frac{1}{2} v t_1 = \frac{1}{2} \left( \frac{\alpha \beta T}{\alpha + \beta} \right) \left( \frac{v}{\alpha} \right) \]
- The area of the second triangle (deceleration phase) is:
\[ s_2 = \frac{1}{2} v t_2 = \frac{1}{2} \left( \frac{\alpha \beta T}{\alpha + \beta} \right) \left( \frac{v}{\beta} \right) \]
- Therefore, the total displacement is:
\[ s = s_1 + s_2 = \frac{1}{2} v t_1 + \frac{1}{2} v t_2 = \frac{1}{2} v (t_1 + t_2) = \frac{1}{2} v T \]
- Substituting the value of \( v \):
\[ s = \frac{1}{2} \left( \frac{\alpha \beta T}{\alpha + \beta} \right) T = \frac{\alpha \beta T^2}{2(\alpha + \beta)} \]

### Final Answer:
\[ s = \frac{\alpha \beta T^2}{2(\alpha + \beta)} \]