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A particle starts from rest and traverses a distance $l$ with uniform acceleration, then moves uniformly over a further distance $2 l$ and finally comes to rest after moving a further distance $3 l$ under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its ways is
$1 / 5$
$2 / 5$
$3 / 5$
$4 / 5$
Solution
(c)
Let $v_m$ be the maximum speed,
$v_m=a_1 t_1 \text { and } v_m=\sqrt{2 a_1} l$
$t_2=\frac{2 l}{v_m} \quad \text { and } \quad v_m=a_2 t_3=\sqrt{2 a_2(3 l)}$
Now, average speed $v_{ av }=\frac{l+2 l+3 l}{t_1+t_2+t_3}$
$v_{ av } =\frac{6 l}{\left(v_m / a_1\right)+\left(2 l / v_m\right)+\left(v_m / a_2\right)}$
$=\frac{6 l}{\left(\frac{v_m}{v_m^2 / 2 l}\right)+\left(\frac{2 l}{v_m}\right)+\left(\frac{v_m}{v_m^2 / 6 l}\right)}$
$=\frac{6 l}{\left(10 l / v_m\right)}=\frac{3 v_m}{5}$
$\frac{v_{ av }}{v_m} =\frac{3}{5}$
