How do you calculate acceleration, retardation, and distance in a train journey?

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Discussion Overview

The discussion revolves around calculating acceleration, retardation, and distance in a train journey, specifically using uniformly accelerated motion equations. Participants are attempting to derive the value of acceleration 'a' based on given conditions and relationships between speed, distance, and time.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant describes a scenario where a train accelerates uniformly from rest to a speed 'v' and then decelerates uniformly to rest, proposing to use the average speed formula to find 'a'.
  • Another participant requests the equations set up by the first poster to assist in the calculations.
  • A participant outlines their approach, calculating times 't1' and 't2' for acceleration and deceleration, respectively, and expresses difficulty in solving the resulting equations.
  • Further elaboration includes the use of the area of a triangle to derive expressions for distance 's' and total time, leading to a relationship between average speed and distance.
  • One participant questions the validity of a derived equation, expressing confusion over the relationship between 's' and 't1 + t2'.
  • Another participant claims to have found a value for 'a' as 16/3 based on their substitutions into the equations.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the equations or the derived values, with some expressing confusion and others proposing different interpretations or calculations.

Contextual Notes

There are unresolved mathematical steps and dependencies on the assumptions made regarding the relationships between speed, distance, and time. Some participants express uncertainty about the correctness of their equations and results.

markosheehan
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a train accelerates uniformly from rest to a speed v m/s with uniform acceleration a m/s^2 it then decelerates uniformly to rest with uniform retardation 3a m/s^2. the total distance traveled is s metres. if the average speed for the whole journey is square root (s/2) find the value of a. does anyone know how to work this out using uvast equations

ive set up equations but i am lost in all of the variables
 
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markosheehan said:
ive set up equations
Then share them.
 
Ok so average speed =total distance/total time
So v=0+at1 so t1 =v/a doing the same for time 2 you get t2= v/3a
Then using area of a triangle .5*v(v/a +v/3a)
Then I put this into my formula above as well as the total time and equal it to ✓s/2 and I have an expression for s.
Anyway when I solve this I got no where.
 
any helping info at all?
 
Sorry for the delay.

So v=0+at1 so t1 =v/a doing the same for time 2 you get t2= v/3a
Then using area of a triangle .5*v(v/a +v/3a)
So $t_1=\dfrac{v}{a}$, $t_2=\dfrac{v}{3a}$; therefore, $t_1+t_2=\dfrac{4v}{3a}$. Also, $s=\dfrac{1}{2}v(t_1+t_2)=\dfrac{2v^2}{3a}$.

The average speed is $\dfrac{s}{t_1+t_2}$, which equals $\sqrt{\dfrac{s}{2}}$ by assumption. Taking squares of both sides, we get $\dfrac{s^2}{(t_1+t_2)^2}=\dfrac{s}{2}$, from where $2s=(t_1+t_2)^2$. Now substitute the expressions for $s$ and $t_1+t_2$. Then $v^2$ should cancel and you get an equation in $a$.
 
Evgeny.Makarov said:
Sorry for the delay.

So $t_1=\dfrac{v}{a}$, $t_2=\dfrac{v}{3a}$; therefore, $t_1+t_2=\dfrac{4v}{3a}$. Also, $s=\dfrac{1}{2}v(t_1+t_2)=\dfrac{2v^2}{3a}$.

The average speed is $\dfrac{s}{t_1+t_2}$, which equals $\sqrt{\dfrac{s}{2}}$ by assumption. Taking squares of both sides, we get $\dfrac{s^2}{(t_1+t_2)^2}=\dfrac{s}{2}$, from where $2s=(t_1+t_2)^2$. Now substitute the expressions for $s$ and $t_1+t_2$. Then $v^2$ should cancel and you get an equation in $a$.

Im sorry I am not sure about some of your answer.
if t1+t2=2v^2/3a and s=2v^2/3a I am not sure how 2s=(t1+t2)^2 as 2(2v^2/3a) does not equal (2v^2/3a)^2

im getting a=16/3 when i sub those in thanks.
 
markosheehan said:
if t1+t2=2v^2/3a
I wrote a different formula.
 

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