Recent content by vryC0nfused

Oh! Okay, I was thinking that I was supposed to calculate the individual average speed for each of the intervals and then somehow add them up. Thank you so much, I see why that wouldn't make any sense now! :)

I think the distance covered of y meters would be larger since you're starting from 100 mph instead of from rest or 0 mph. I incorrectly assumed that I was starting with a uniform velocity of 28 m/s when I determined the distance traveled.

Ok, since the train is starting from "a dead stop" and I'm assuming that the acceleration is constant, I used the equation x = x_0 + (1/2)(V_{x0} + V_x)t to find the distance. x = (0 m) + (1/2)(0 m/s + 28 m/s)(120 s) ---> x = 1,680 meters I think this is the correct distance covered during this...

Oh! I think that's where I'm making my mistake, I'm treating the train as if it's moving at a constant velocity! I'll check my notes for the equation I need to use.

Wouldn't the average speed for the first 120 second interval also be 28 m/s? You'd divide the total distance (3,360 meters) by the total time taken to cover that distance (120 seconds) and get 28 m/s.

Oh! I'll make sure to provide evidence of my attempts in future questions! A1. During the first 2 minutes (120 seconds) the train covered a distance of 3,360 meters. (28 m/s) (120 s) = 3,360 meters A2. Using one of of the famous five ( V_x = V_{x0} + a_xt ) and treating the train like it's...

Okay, that sounds like a good idea.

I think it's average over distance.

TL;DR Summary: A train on a level stretch of tracks, starting from a dead stop, steadily speeds up to 28 m/s in 2.0 minutes, and then slows to 18 m/s with a constant deceleration of 0.011 m/s^2. What is the average speed of the train for this whole thing (speeding up and slowing down)? I know...