Find the total distance taken by the train.

[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.

 

[SteamKing]

[ 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]

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]

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