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- anonymous

help....
a uniformly accelerating train passes a person on the ground.If the speeds of the points 1, 2 & 3 on the train as they pass the person are u1, u2 & u3 respectively and u1/u2 = u2/u3, then s1/s2 is (where s1 is the distance between 1 & 2 ; s2 between 2 & 3 respectively)
a. s1/s2 = u2/u3
b. s1/s2 =(u1/u2)^2
pls tell me how 2 do this....

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- anonymous

- chestercat

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- anonymous

Consider the Newton's third equation of motion: v^2+u^2=2as,
where "v" denotes final velocity, and "u" denotes initial velocity.
Here, acceleration is uniform, therefore, "2a" is constant.
Hence, v^2-u^2 is proportional to "s"......(i)
Consider the first case. Here, "u2" is the final velocity, "u1" is the initial velocity and "s1" is the distance between the points "1" and "2".
Therefore, (u2)^2-(u1)^2=s1.....(ii)
But given that u1/u2=u2/u3, or (u2)^2=u1u3
Hence, by substitution, we get equation(ii) as follows:
u1u3-(u1)^2=s1
Or, u1(u3-u1)=s1..........(iii)
Consider the second case. Here, (u2) is the initial velocity, and (u3) is the final velocity, and "s2" is the distance between points "2" and "3".
So, we get, (u3)^2-(u2)^2=s2
But we know that (u2)^2=u1u3
Therefore, (u3)^2- u1u3=s2
Or, u3(u3-u1)=s2
Now, on solving, we get, s1/s2=[u1(u3-u1)]/[u3(u3-u1)]=u1/u3
We know that u1u3=(u2)^2
Or, (u1u3)/(u3u2)=(u2*u2)/(u3u2)
Or, (u1/u3)(u3/u2)=u2/u3
Or, u1/u2=u2/u3
Therefore, the answer = u2/u3

- anonymous

thank u...! :)

- anonymous

I don't get the last step.

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