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

AI Thread Summary
A train accelerates uniformly from rest to speed v m/s with acceleration a m/s² and then decelerates to rest with retardation 3a m/s². The total distance traveled is s meters, and the average speed for the journey is √(s/2). The discussion involves setting up equations for time and distance, leading to expressions for total time and average speed. Participants derive relationships between distance, time, and acceleration, ultimately aiming to solve for a. The calculations suggest that substituting the derived values leads to a final result of a = 16/3.
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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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