Welcome to our community

Be a part of something great, join today!

Physics journey

markosheehan

Member
Jun 6, 2016
136
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 travelled 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
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,533

markosheehan

Member
Jun 6, 2016
136
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 aswell 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.
 

markosheehan

Member
Jun 6, 2016
136
any helping info at all?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,533
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 $a$.
 

markosheehan

Member
Jun 6, 2016
136
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 im not sure about some of your answer.
if t1+t2=2v^2/3a and s=2v^2/3a im 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.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,533