To solve the problem, let's break down the given information and apply the equations of motion. Here's a summary of the problem:

• The train starts at point A with an initial speed of $110 \, \text{km/h}$.
• It decelerates to a speed $v \, \text{km/h}$ in $t$ minutes.
• After reaching $v$, it travels at a constant speed of $v \, \text{km/h}$ for some time.
• Then, it accelerates for $3t$ minutes and reaches the speed of $110 \, \text{km/h}$ at point B.
• The distance traveled during the constant velocity phase is $2.5 \, \text{km}$.
• The total distance from A to B is $6.5 \, \text{km}$.
• The total time taken to travel from A to B is 6 minutes.

We need to show that the equation $v(6 - 4t) = 150$ holds.

Step 1: Analyze the deceleration phase

• Initial speed: $110 \, \text{km/h}$
• Final speed: $v \, \text{km/h}$
• Time taken: $t$ minutes, which is $\frac{t}{60} \, \text{hours}$

Let the deceleration be $a_1$. Using the first equation of motion: $v = 110 - a_1 \cdot \frac{t}{60}$ Thus, $a_1 = \frac{110 - v}{t/60} = \frac{60(110 - v)}{t}$.

Step 2: Analyze the constant velocity phase

• Speed: $v \, \text{km/h}$
• Distance: $2.5 \, \text{km}$

Time taken during this phase, $t_2$, is given by: $t_2 = \frac{2.5}{v} \, \text{hours}$ Convert $t_2$ to minutes: $t_2 = \frac{2.5 \times 60}{v} \, \text{minutes} = \frac{150}{v} \, \text{minutes}$

Step 3: Analyze the acceleration phase

• Initial speed: $v \, \text{km/h}$
• Final speed: $110 \, \text{km/h}$
• Time taken: $3t$ minutes, which is $\frac{3t}{60} \, \text{hours}$

Let the acceleration be $a_2$. Using the first equation of motion: $110 = v + a_2 \cdot \frac{3t}{60}$ Thus, $a_2 = \frac{60(110 - v)}{3t} = \frac{20(110 - v)}{t}$.

Step 4: Total time and total distance

Total time $T$ is given as 6 minutes: $t + t_2 + 3t = 6$ Substitute $t_2 = \frac{150}{v}$: $t + \frac{150}{v} + 3t = 6$ Simplify: $4t + \frac{150}{v} = 6$ Multiply the entire equation by $v$ to get rid of the fraction: $4tv + 150 = 6v$ Rearrange to show the required equation: $6v - 4tv = 150$ Factor out $v$ from the left side: $v(6 - 4t) = 150$

Conclusion

We have shown that $v(6 - 4t) = 150$ as required.

Would you like any further details or have any questions?

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Here are some related questions to consider:

1. What would the equation be if the total distance AB were different?
2. How would the solution change if the time taken to decelerate was different?
3. How would this problem change if the train accelerated uniformly from the start instead of decelerating?
4. Can you derive the equation for $v$ if the train's acceleration phase were to take $4t$ minutes instead of $3t$?
5. What other scenarios can be analyzed using the first equation of motion?

Tip: When dealing with problems that involve multiple phases of motion (deceleration, constant velocity, acceleration), it's useful to break down the problem into sections and apply the relevant equations of motion to each phase.