# 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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