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. $\frac{4}{5}$
• B. $\frac{3}{5}$
• C. $\frac{2}{5}$
• D. $\frac{1}{5}$

Answer 1

Answer: (B)

Let $t_{1}, t_{2}$ and $t_{3}$ 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.

The particle moves with uniform acceleration from $A$ to $B$. For motion from $A$ to $B$
Average velocity $=\frac{0+v}{2}=\frac{v}{2}$
Time taken, $t_{1}=\frac{2 x}{v / 2}=\frac{4 x}{v}$
The particle moves with uniform velocity from $B$ to $C$,
$\therefore t_{2}=\frac{4 x}{v}$
The particle moves with uniform retardation from $C$ to $D$,
$\therefore t_{3}=\frac{6 x}{(0+v) / 2}=\frac{12 x}{v}$
Total time taken $=t_{1}+t_{2}+t_{3}$
$=\frac{4 x}{v}+\frac{4 x}{v}+\frac{12 x}{v}=\frac{20 x}{v}$
$v_{ av }=\frac{2 x+4 x+6 x}{20 x / v}=\frac{12 v}{20} $or
$ \frac{v_{ av }}{v}=\frac{12}{20}=\frac{3}{5}$