Question

A particle starts from rest and traverses a distance $2x$ with uniform acceleration, then moves uniformly over a further distance $4x$ and finally comes to rest after moving a further distance $6x$ 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 time taken by the particles to cover the distance $2x$ , $4x$ and $6x$ 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, $t_{1}=\frac{2 x}{v/2}$
Particle moves with uniform retardation from $C$ TO $D$ , $V_{a v g}=\frac{0 + v}{2}=v/2$
Time taken , $t_{3}=\frac{6 x}{\left(\right. v/2 \left.\right)}$
Total time = $t_{1}+t_{2}+t_{3}$
$t=\frac{2 x}{v/2}+\frac{4 x}{v}+\frac{6 x}{v/2}=\frac{20 x}{v}$
Average Velocity over= $\frac{6 x + 2 x + 4 x}{20 x/v}=\frac{3}{5}v$
Required Ratio = $\frac{Average Speed}{v}=\frac{3}{5}$