Train Velocity-Time Graph: Uniform Acceleration and Retardation Motion Explained

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Discussion Overview

The discussion revolves around the motion of a train characterized by uniform acceleration, constant velocity, and uniform retardation. Participants explore the mathematical representation of this motion through a velocity-time graph and seek to understand the relationship between the graph's shape and the total distance traveled by the train.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant describes the train's motion, detailing the phases of acceleration, constant velocity, and retardation.
  • Another participant questions how velocity varies mathematically under uniform acceleration, prompting a discussion about the relevant formula.
  • Several participants express uncertainty about the graphing process and the calculation of distance from the graph.
  • There is a suggestion that the graph resembles a trapezium, leading to a discussion about the area representing total distance.

Areas of Agreement / Disagreement

Participants generally agree on the use of the trapezium area to calculate total distance, but there is uncertainty regarding the graphing process and the application of the formula for velocity.

Contextual Notes

Some participants do not provide answers to specific questions, indicating potential gaps in understanding or knowledge about the graphing and calculation processes.

Who May Find This Useful

Individuals interested in kinematics, particularly those looking to understand the graphical representation of motion and the relationship between acceleration, velocity, and distance.

Jerome1
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A train starts from rest from a station and travels with uniform acceleration 0.5m/s^2 for 20s. it travels with uniform velocity for another 30s, the brakes are then applied so that a uniform retardation is obtained and the train comes to rest in a further 10s. sketch the velocity-time graph of this motion. Using your graph, calculate the total distance traveled by the train

please can i see the graph, and also include explanation so that i can understand better
 
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First, let me ask you if the velocity changes uniformly, that is, the acceleration is constant, how will the velocity vary, mathematically speaking?
 
i don't have answers to that.
 
Welcome to MHB, Jerome! :)

Jerome said:
A train starts from rest from a station and travels with uniform acceleration 0.5m/s^2 for 20s. it travels with uniform velocity for another 30s, the brakes are then applied so that a uniform retardation is obtained and the train comes to rest in a further 10s. sketch the velocity-time graph of this motion. Using your graph, calculate the total distance traveled by the train

please can i see the graph, and also include explanation so that i can understand better

Jerome said:
i don't have answers to that.

The formula for velocity with uniform acceleration is:
$$v = v_0 + a t$$
where $v$ is the speed at time $t$, $v_0$ is the initial speed, and $a$ is the uniform acceleration.

Is this formula known to you?

Anyway, in your problem, we can deduce that $v_0 = 0$, since the train is initially at rest.
And $a=0.5 m/s^2$ is given.

Do you know how to draw the graph of $v=0.5 t$ for the time period of 0 to 20s?
What will be the speed at $t=20s$?
 
I like Serena said:
Welcome to MHB, Jerome! :)The formula for velocity with uniform acceleration is:
$$v = v_0 + a t$$
where $v$ is the speed at time $t$, $v_0$ is the initial speed, and $a$ is the uniform acceleration.

Is this formula known to you?

Anyway, in your problem, we can deduce that $v_0 = 0$, since the train is initially at rest.
And $a=0.5 m/s^2$ is given.

Do you know how to draw the graph of $v=0.5 t$ for the time period of 0 to 20s?
What will be the speed at $t=20s$?
i got that thanks, the graph looks like a trapezium, so the area of the trapezium is the total distance right?
 
Jerome said:
i got that thanks, the graph looks like a trapezium, so the area of the trapezium is the total distance right?

Yep!
 

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