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AIIMS 201826 May
Exam
AIIMS
Year
2018
Date
26 May
Subject
Physics
Chapter
Motion in One Dimension

AIIMS 2018 · 26 May · Physics · Motion in One Dimension

A body starts from rest and travels a distance S with uniform...

A body starts from rest and travels a distance SS with uniform acceleration, then moves uniformly a distance 2S2 S and finally comes to rest after moving further 5S5 S under uniform retardation. The ratio of the average velocity to maximum velocity is

Options

  1. A25\frac{2}{5}
  2. B35\frac{3}{5}
  3. C47\frac{4}{7}
  4. D57\frac{5}{7}

Correct answer

C. 47\frac{4}{7}

Step-by-step solution

Graphically, the area of v−tv-t curve represents displacement: S=12vmax⁡t1S=\frac{1}{2} v_{\max } t_1 or t1=2Svmax⁡2S=vmax⁡t2ort2=2Svmax⁡5S=12vmax⁡t3ort3=10Svmax⁡t_1=\frac{2 S}{v_{\max }} 2 S=v_{\max } t_2 or t_2=\frac{2 S}{v_{\max }} 5 S=\frac{1}{2} v_{\max } t_3 or t_3=\frac{10 S}{v_{\max }}vav= Total displacement  Total time =S+2S+5S2Svmax+2Svmax⁡+10Svmax⁡vavvmax=8S14S=47\begin{aligned} v_{\mathrm{av}} & =\frac{\text{ Total displacement }}{\text{ Total time }} \\ & =\frac{S+2 S+5 S}{\frac{2 S}{v_{\mathrm{max}}}+\frac{2 S}{v_{\max }}+\frac{10 S}{v_{\max }}} \\ \frac{v_{\mathrm{av}}}{v_{\mathrm{max}}} & =\frac{8 S}{14 S}=\frac{4}{7}\end{aligned}$$
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