# Find the total distance taken by the train.

• Electgineer
In summary, the train starts from rest and accelerates at a rate of 6m/s² for 20 seconds to attain a constant speed. It then travels at this constant speed for 20 seconds before decelerating at a rate of 3m/s² for another 20 seconds. To calculate the total distance traveled by the train, we can use the SUVAT equations for each type of motion and add up the distances. This can be done by breaking the problem into three parts: calculating the distance and velocity after acceleration, calculating the distance after traveling at constant velocity, and calculating the additional distance and final velocity after deceleration.
Electgineer
[ A train starting from rest accelerate at the rate of 6m/s² for 20 seconds to attain a constant speed and it traveled for another 20 seconds and decelerate at the rate of 3m/s² for 20 seconds. Calculate the total distance (in kilometer) traveled by the train.
That's the question. I found it confusing when I was given the deceleration value.

Electgineer said:
[ A train starting from rest accelerate at the rate of 6m/s² for 20 seconds to attain a constant speed and it traveled for another 20 seconds and decelerate at the rate of 3m/s² for 20 seconds. Calculate the total distance (in kilometer) traveled by the train.
That's the question. I found it confusing when I was given the deceleration value.
Why? What's confusing about the deceleration of the train?

The train starts from rest and accelerates to a constant speed. The train travels at this constant speed for a certain time interval. The train then slows down for another time period.

Apply the SUVAT equations for each type of motion and add up the distances traveled.

In both time intervals the speed of the train is changing.
In the first 20 s it is increasing.
In the next 20 s it is decreasing.
You can use the constant acceleration equations to calculate this.
How would you change the equations to calculate for the deceleration?

andrevdh said:
In both time intervals the speed of the train is changing.
In the first 20 s it is increasing.
In the next 20 s it is decreasing.
You can use the constant acceleration equations to calculate this.
How would you change the equations to calculate for the deceleration?
Try doing the problem in 3 parts. Part 1: What is the distance and velocity after accelerating at 6 m/s^2 for 20 seconds? Part 2: What is the distance after traveling at the constant velocity from Part 1 for an additional 20 seconds? Part 3: What is the additional distance and the final velocity after accelerating -3 m/s^2 for an additional 20 seconds?

Chet

## What is the total distance taken by the train?

The total distance taken by the train is the sum of all the distances traveled during its journey, including any stops or detours along the way.

## How is the total distance calculated for a train journey?

The total distance is calculated by adding up the distances between each stop on the train's route. This can also include any additional distance traveled due to detours or delays.

## Is the total distance the same for every train journey?

No, the total distance can vary for each train journey depending on the route, stops, and any unforeseen circumstances that may affect the train's travel time.

## Can the total distance be affected by factors other than the train's route?

Yes, the total distance can also be affected by factors such as weather conditions, speed restrictions, and maintenance work on the train tracks.

## Why is knowing the total distance taken by the train important?

Knowing the total distance taken by the train can help with planning and scheduling future train journeys, as well as determining the efficiency and reliability of the train service.

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