Question

Train stops at two stations 'd' dist. apert.

Time taken = t$\checkmark$ Crux: Never uniform velo

acc = $\alpha\checkmark$

retar= B$\checkmark$ Find reln in ($\alpha$,B,t,d):

Answer 1

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{t^2}{2d}$

Answer 2

The problem describes a train's motion between two stations, starting from rest and ending at rest. The motion consists of two phases: uniform acceleration and uniform retardation, with no period of constant velocity.

Let:

• $d$ be the total distance between the stations.
• $t$ be the total time taken for the journey.
• $\alpha$ be the acceleration.
• $\beta$ be the retardation (deceleration).
• $v_{max}$ be the maximum velocity attained by the train.
• $t_1$ be the time taken during acceleration.
• $t_2$ be the time taken during retardation.

1. Velocity-Time Relationship:

The train starts from rest, accelerates to $v_{max}$, and then decelerates to rest. The velocity-time graph for this motion is a triangle.

• During acceleration: $v_{max} = \alpha t_1 \implies t_1 = \frac{v_{max}}{\alpha}$
• During retardation: $0 = v_{max} - \beta t_2 \implies v_{max} = \beta t_2 \implies t_2 = \frac{v_{max}}{\beta}$

The total time $t$ is the sum of $t_1$ and $t_2$: $t = t_1 + t_2 = \frac{v_{max}}{\alpha} + \frac{v_{max}}{\beta}$ $t = v_{max} \left( \frac{1}{\alpha} + \frac{1}{\beta} \right)$ (Equation 1)

2. Distance-Time Relationship:

The total distance $d$ is the area under the velocity-time graph (a triangle). Area = $\frac{1}{2} \times \text{base} \times \text{height}$ $d = \frac{1}{2} \times t \times v_{max}$ From this, we can express $v_{max}$ in terms of $d$ and $t$: $v_{max} = \frac{2d}{t}$ (Equation 2)

3. Combining the Relationships:

Substitute the expression for $v_{max}$ from Equation 2 into Equation 1: $t = \left( \frac{2d}{t} \right) \left( \frac{1}{\alpha} + \frac{1}{\beta} \right)$ Multiply both sides by $t$: $t^2 = 2d \left( \frac{1}{\alpha} + \frac{1}{\beta} \right)$ Rearrange the equation to find the desired relation: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{t^2}{2d}$

This is the relationship between $\alpha$, $\beta$, $t$, and $d$.

Explanation of the solution:

The problem involves two phases of motion: uniform acceleration ($\alpha$) from rest to maximum velocity ($v_{max}$), and uniform retardation ($\beta$) from $v_{max}$ back to rest. The total time is $t$ and total distance is $d$.

1. Express the maximum velocity $v_{max}$ in terms of acceleration and time ($t_1$) for the first phase, and in terms of retardation and time ($t_2$) for the second phase.
2. Sum $t_1$ and $t_2$ to get the total time $t$ in terms of $v_{max}$, $\alpha$, and $\beta$.
3. Recognize that the velocity-time graph is a triangle. The area under this graph represents the total distance $d$. Use $d = \frac{1}{2} v_{max} t$ to express $v_{max}$ in terms of $d$ and $t$.
4. Substitute the expression for $v_{max}$ from step 3 into the equation from step 2 and simplify to obtain the final relation.

Answer:

The relation between $\alpha$, $\beta$, $t$, and $d$ is: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{t^2}{2d}$