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

Click For Summary
The discussion focuses on a train's motion characterized by uniform acceleration, constant velocity, and uniform retardation. The train accelerates at 0.5 m/s² for 20 seconds, travels at a constant speed for 30 seconds, and then decelerates to a stop over 10 seconds. Participants discuss how to sketch the corresponding velocity-time graph, which resembles a trapezium, and calculate the total distance traveled using the area of this trapezium. The formula for velocity during uniform acceleration is highlighted, emphasizing the relationship between time, acceleration, and velocity. Understanding these concepts is essential for visualizing and calculating the train's motion accurately.
Jerome1
Messages
17
Reaction score
0
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
 
Mathematics news on Phys.org
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!