Question

A particle starts from rest and traverses a distance $l$ with uniform acceleration, then moves uniformly over a further distance $2l$ and finally comes to rest after moving a further distance $3l$ 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. $\frac{11}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is C $\frac{3}{5}$

$AB=L,BC=2L,CD=3L$

It starts from rest and accelerates and then remains in uniform velocity and then retards, this means the maximum velocity will be at point B and let it be v

The area under velocity time graph is displacement and hence we have three equations corresponding to regions $AB,BC$ and $CD$

$\frac{v{t}_1}{2}=2,v{t}_2=2L,v{t}_3=3L$

$t_1+t_2+t_3=+\frac{2L}{v}+\frac{2L}{v}+\frac{6L}{v}=\frac{10L}{v}$

and the average velocity

$=\frac{totaldisplacement}{totaltime}$

$\overline{v}=\frac{6L}{t_1+t_2+t_3}=\frac{3v}{5}$

Hence, $\frac{\overline{v}}{v}=\frac{3}{5}$