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Grade 11Mechanics

A train is uniformly accelerated for a distance equal to 1/m of the distance between two stations and uniformly retarded for 1/n of the distance b/w thesthese stations .find the ratio of maximum to the average velocity during the journey.
Thank for the previous answer!

Simran

8 Years agoGrade 11

1 Answer

Eshan

8 Years ago

Dear student,

Let the maximum velocity attained be $v_0$ and acceleration for distance $\dfrac{s}{m}$ be $a_1$ , and that for distance $\dfrac{s}{n}$ be $a_2$ .
Hence we have $\dfrac{s}{ m}=\dfrac{v_0^2}{2a_1}$ and $\dfrac{s}{ n}=\dfrac{v_0^2}{2a_2}$

Also the total time taken in the motion is sum of those periods with acceleration, constant velocity and retardation= $t_1+t_2+t_3=\dfrac{v_0}{a_1}+\dfrac{s(1-\dfrac{1}{m}-\dfrac{1}{n})}{v_0}+\dfrac{v_0}{a_2}$

Thus average velocity= $\dfrac{s}{t}$

Hence the required ratio= $\dfrac{v_0}{s/t}=\dfrac{v_0}{s}t=\dfrac{v_0^2}{a_1s}+(1-\dfrac{1}{m}-\dfrac{1}{n})+\dfrac{v_0^2}{a_2s}$

$=\dfrac{2}{m}+(1-\dfrac{1}{m}-\dfrac{1}{n})+\dfrac{2}{n}=1+\dfrac{1}{m}+\dfrac{1}{n}$

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