# Find the total distance taken by the train.

[ 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.

Related Introductory Physics Homework Help News on Phys.org
SteamKing
Staff Emeritus
Homework Helper
[ 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.

andrevdh
Homework Helper
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?

Chestermiller
Mentor
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