If a body starting from rest moves with acceleration alpha for certain time and then decceleration at the rate beta.until it stopa and `t` is the total time of its motion find:-1)maximum velocity 2) average velocity 3)distance travelled by the body

To tackle this problem, we need to break it down into two distinct phases: the acceleration phase and the deceleration phase. Let's denote the time of acceleration as \( t_1 \) and the time of deceleration as \( t_2 \). The total time \( t \) is then the sum of these two phases: \( t = t_1 + t_2 \).

1. Maximum Velocity

During the acceleration phase, the body starts from rest and accelerates at a rate of \( \alpha \). The formula for velocity when starting from rest is:

Velocity (v) = Initial Velocity + Acceleration × Time

Since the initial velocity is zero, the maximum velocity \( v_{max} \) at the end of the acceleration phase can be expressed as:

v_{max} = \alpha \cdot t_1

2. Average Velocity

The average velocity \( v_{avg} \) over the entire motion can be calculated using the total distance traveled divided by the total time. However, we first need to find the distance traveled during both phases.

Distance During Acceleration

The distance \( d_1 \) traveled during the acceleration phase can be calculated using the formula:

d_1 = Initial Velocity × Time + 0.5 × Acceleration × Time²

Substituting the values, we get:

d_1 = 0 + 0.5 × \alpha × t_1² = 0.5 \alpha t_1²

Distance During Deceleration

For the deceleration phase, the body starts with the maximum velocity \( v_{max} \) and decelerates at a rate of \( \beta \). The time taken to stop is \( t_2 \), and the distance \( d_2 \) can be calculated as:

d_2 = v_{max} × t_2 - 0.5 × \beta × t_2²

Substituting \( v_{max} \) gives:

d_2 = (\alpha t_1) t_2 - 0.5 \beta t_2²

3. Total Distance and Average Velocity

The total distance \( d \) traveled by the body is the sum of \( d_1 \) and \( d_2 \):

d = d_1 + d_2 = 0.5 \alpha t_1² + (\alpha t_1) t_2 - 0.5 \beta t_2²

Now, the average velocity \( v_{avg} \) can be calculated as:

v_{avg} = \frac{Total Distance}{Total Time} = \frac{d}{t} = \frac{0.5 \alpha t_1² + (\alpha t_1) t_2 - 0.5 \beta t_2²}{t_1 + t_2}

Summary of Results

• Maximum Velocity: v_{max} = α t_1
• Average Velocity: v_{avg} = (0.5 α t_1² + α t_1 t_2 - 0.5 β t_2²) / (t_1 + t_2)
• Total Distance: d = 0.5 α t_1² + (α t_1) t_2 - 0.5 β t_2²

By following these steps, you can effectively analyze the motion of the body under the given conditions. If you have any further questions or need clarification on any part, feel free to ask!