Question

A particle stars from rest covers a distance of x with constant acceleration then move with constant velocity and cover distance 2x then after come to rest will constant retardation with cover distance of 3x, then ratio of $V_{max}/V_{avg}$ will be

• A. 3:5
• B. 5:3
• C. 7:5
• D. 5:7

Answer 1

The correct option is B 5:3

Suppose the constant acceleration in the first part of the motion is $a$ then time taken in covering the distance $x$ and achieving velocity $v$ will be $t_1=\sqrt{\frac{2x}{a}}$

and this maximum velocity will be $V_{max}=a{t}_1=\sqrt{2ax}$

so time taken taken in the second part of the motion i.e. in covering next $2x$ will be $t_2=\frac{2x}{V_{max}}=\frac{2x}{\sqrt{2ax}}=\sqrt{\frac{2x}{a}}$

Now the distance covered is $3x$ before coming to rest so the retardation will be $a_1=\frac{V_{max}^2}{2S}=\frac{2ax}{2\times 3x}=\frac{a}{3}$

so time taken in covering the last segment $\text{when the velocity goes to zero from its max value}$

is $t_3=\frac{V_{max}}{a_1}=3\sqrt{\frac{2x}{a}}$

so the total time of the motion will be $t_1+t_2+t_3=5\sqrt{\frac{2x}{a}}$

and the toatl distance of the motion is $x+2x+3x=6x$

so $V_{avg}=\frac{6x}{time}=\frac{6}{5}\sqrt{\frac{ax}{2}}$

and as derived earlier the $V_{max}=\sqrt{2ax}$

the ratio is $\frac{V_{max}}{V_{avg}}=\frac{5}{3}$