# Particle accelerating

### Particle accelerating

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

markosheehan

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### Re: Particle accelerating

Split the journey into two legs. In the first leg we are accelerating, in the second we are decelerating.
To avoid confusion I will use capital letters for the general formulas, and small letters for the problem variables.
$U$ = initial velocity
$V$ = final velocity
$A$ = acceleration
$S$ = distance
$T$ = time taken

First leg:
Information we are given: $U=0, V=v, A=a$
We need to calculate the distance $S$ and time taken $T$, which we'll call $s_1$, and $t_1$.
$V=U+AT$ implies $v=at_1$, so $t_1=v/a$.
$V^2=U^2+2AS$ implies $v^2=2as_1$, so $s_1=v^2/(2a)$.

Second leg:
Information we are given: $U=v, V=0, A=-3a$
We need to calculate the distance $S$ and time taken $T$, which we'll call $s_2$, and $t_2$.
$V=U+AT$ implies $0=v-3at_2$, so $t_2=v/(3a)$.
$V^2=U^2+2AS$ implies $0 = v^2-6as_2$, so $s_2=v^2/(6a)$.

Putting it all together:
We are told in the question that $s=s_1+s_2$, and that $\sqrt{s/2}=(s_1+s_2)/(t_1+t_2)$,
so $s=v^2/(2a)+v^2/(6a) = 2v^2/(3a)$,
$\sqrt{v^2/(3a)} = (2v^2/(3a))/(v/a+v/(3a))$,
$\sqrt{v^2/(3a)} = (2v^2/(3a))/(4v/(3a))$,
$\sqrt{v^2/(3a)} = v/2$,
$v^2/(3a) = v^2/4$,
$3a = 4$,
$a=4/3$

Hope this helped,

R. Baber.

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