Question

a uniformly accelerating train passes a person on the ground. if the speed of the points 1, 2, abd 3 on the train as they passes the person are u1 u2 u3 respectively and u1/u2=u2/u3 then s1/s2 is ?( where s1 is the distance between 1 and 2 s2 between 2 and 3 respectively

Answer 1

generically, the train can be modeled as such
\[a(t) = a\]\[v(t) = at+v\]\[s(t)=\frac{1}{2}at^2+vt+s\]

Answer 2

we are given that:\[\frac{a(t_1)+v}{{a(t_2)+v}}=\frac{a(t_2)+v}{{a(t_3)+v}}\]

we are given that:\[(a(t_1)+v)(a(t_3)+v)=(a(t_2)+v)^2\]
\[a^2(t_1)(t_3)+av(t_1)+av(t_3)+v^2=a^2(t_2)^2+2av(t_2)+v^2\]
\[a^2(t_1)(t_3)+av(t_1)+av(t_3)=a^2(t_2)^2+2av(t_2)\]
\[a(t_1)(t_3)+v(t_1+t_3) =a(t_2)^2+2v(t_2)\]

hmmm, still trying to iron out my idea :)

Answer 3

I think using this relationship might be more fruitful:\[v^2=u^2+2as\]therefore:\[2as=v^2-u^2\]so we get:\[2as_1=u_2^2-u_1^2\]and:\[2as_2=u_3^2-u_2^2\]you should be able to solve from here...

Answer 4

if i can assume that at v(0) the train has a speed of 0, then all my v parts go zero ....

asnaseers method should prove useful, but im a glutton for punishment and like to brute math my way thru ;)

Answer 5

lol!