A particle starts from rest and travels a distance s with uniform acceleration, then it travels a distance 2s with uniform speed finally it travels a distance 3s with uniform retardation and comes to rest. If the complete motion of the particle is a straight line then the ratio of its average velocity to maximum velocity is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

6/7

3/5

2/5

4/5

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Maximum velocity $v_{max} = \sqrt{2 as}$
$\begin{array}{l} v_{avg} = \frac{6 s}{T} T = t_{1} + t_{2} + t_{3} \\ t_{1} = \sqrt{\frac{2 s}{a}}, t_{2} = \frac{2 s}{v_{max}} = \sqrt{\frac{2 s}{a}} \\ Retardation \Rightarrow B \\ β = \frac{v_{max}^{2}}{6 s} \\ β = \frac{2 a \times s}{6 \times s} = \frac{a}{3} \\ t_{3} = \frac{v_{max}}{β} = 3 \sqrt{\frac{2 a \times s}{a}} = 3 \sqrt{\frac{2 s}{a}} \\ T = 5 \sqrt{\frac{2 s}{a}} \begin{array}{l} v_{avg} = \frac{6}{5} \frac{5}{\sqrt{\frac{2 s}{a}}} \\ v_{avg} = \frac{6}{5} \sqrt{\frac{as}{2}} \\ \frac{v_{avg}}{v_{max}} = \frac{6}{5} \sqrt{\frac{as}{2}} \times \frac{1}{\sqrt{2 as}} = \frac{3}{5} \end{array} \end{array}$