Why? What's confusing about the deceleration of the train?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.
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?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?
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.
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.
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.
Yes, the total distance can also be affected by factors such as weather conditions, speed restrictions, and maintenance work on the train tracks.
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.