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That's the question. I found it confusing when I was given the deceleration value.

- Thread starter Electgineer
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- #1

- 9

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That's the question. I found it confusing when I was given the deceleration value.

- #2

SteamKing

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Why? What's confusing about the deceleration of the train?

That's the question. I found it confusing when I was given the deceleration value.

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.

- #3

andrevdh

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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?

- #4

Chestermiller

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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?

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?

Chet

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