Train Journey: Acceleration, Retardation & Distance Calculations

In summary, the average speed is $\sqrt{\dfrac{s}{2}}$, and the total distance traveled is $2s=(t_1+t_2)^2$.
  • #1
markosheehan
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0
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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  • #2
markosheehan said:
ive set up equations
Then share them.
 
  • #3
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.
 
  • #4
any helping info at all?
 
  • #5
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 $a$.
 
  • #6
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.
 
  • #7
markosheehan said:
if t1+t2=2v^2/3a
I wrote a different formula.
 

FAQ: Train Journey: Acceleration, Retardation & Distance Calculations

What is acceleration?

Acceleration is the rate at which an object's velocity changes over time. It can be calculated by dividing the change in velocity by the change in time.

How is acceleration related to train journeys?

Acceleration is an important factor to consider in train journeys because it affects how quickly the train is able to increase or decrease its speed. This is especially important when the train is stopping or starting at a station.

What is retardation in relation to train journeys?

Retardation is the opposite of acceleration and refers to the rate at which the train is slowing down. It is also known as deceleration and is measured in the same units as acceleration.

How are acceleration and retardation calculated in train journeys?

Acceleration and retardation can be calculated using the formula a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time. The same formula can be used for retardation, but the value of vf will be smaller than vi.

How is distance calculated in train journeys?

Distance can be calculated using the formula d = vi*t + 1/2*a*t^2, where d is distance, vi is initial velocity, t is time, and a is acceleration. This formula takes into account both the initial velocity and the acceleration of the train to determine the distance it has traveled over a given time period.

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