A particle starts from rest and traverses a distance 2 x with uniform acceleration, then moves uniformly over a further distance 4 x and finally comes to rest after moving a further distance 6 x under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its way is

Question

A particle starts from rest and traverses a distance 2 x with uniform acceleration, then moves uniformly over a further distance 4 x and finally comes to rest after moving a further distance 6 x under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its way is (A) (4/5) (B) (3/5) (C) (2/5) (D) (1/5)

Tardigrade Admin · Accepted Answer

Let t1, t2 and t3 be the times taken by the particle to cover the distances 2 x, 4 x and 6 x respectively. Let v be the velocity of the particle at B i.e. maximum velocity. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4f4b14f130e1c539c189be932fa44c1a-.png" /> The particle moves with uniform acceleration from A to B. For motion from A to B Average velocity =(0+v/2)=(v/2) Time taken, t1=(2 x/v / 2)=(4 x/v) The particle moves with uniform velocity from B to C, ∴ t2=(4 x/v) The particle moves with uniform retardation from C to D, ∴ t3=(6 x/(0+v) / 2)=(12 x/v) Total time taken =t1+t2+t3 =(4 x/v)+(4 x/v)+(12 x/v)=(20 x/v) v av =(2 x+4 x+6 x/20 x / v)=(12 v/20) or (v av /v)=(12/20)=(3/5)

$\frac{4}{5}$

$\frac{3}{5}$

$\frac{2}{5}$

$\frac{1}{5}$

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$\frac{4}{5}$

$\frac{3}{5}$

$\frac{2}{5}$

$\frac{1}{5}$