Particle accelerating

Particle accelerating

Postby markosheehan » Fri Dec 09, 2016 8:40 am

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





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Re: Particle accelerating

Postby Guest » Sat Dec 10, 2016 6:15 am

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.
[tex]U[/tex] = initial velocity
[tex]V[/tex] = final velocity
[tex]A[/tex] = acceleration
[tex]S[/tex] = distance
[tex]T[/tex] = time taken

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

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

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

Hope this helped,

R. Baber.

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