Kinematics

JEE Main level, single correct, medium

76 of 308

Question 76 of 308

A body travels a distance ss in tt seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration f\mathrm{f} and in the second part with constant retardation r\mathrm{r}. The value of t\mathrm{t} is given by

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Answer: (A) 2s(1f+1r)\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}

Portion OA, OB corresponds to motion with acceleration ' f\mathrm{f} ' and retardation ' rr ' respectively.
Area of △OAB=S\triangle O A B=S and OB=tO B=t. Let OL=t1O L=t_1,
LB=t2\mathrm{LB}=\mathrm{t}_2 and AL=v,S=12OB⋅AL=12t⋅v;v=2 St\mathrm{AL}=\mathrm{v}, \mathrm{S}=\frac{1}{2} \mathrm{OB} \cdot \mathrm{AL}=\frac{1}{2} \mathrm{t} \cdot \mathrm{v} ; \mathrm{v}=\frac{2 \mathrm{~S}}{\mathrm{t}}
Also, f=vt1,t1=vf=2 stf\mathrm{f}=\frac{\mathrm{v}}{\mathrm{t}_1}, \mathrm{t}_1=\frac{\mathrm{v}}{\mathrm{f}}=\frac{2 \mathrm{~s}}{\mathrm{tf}} and r=vt2,t2=vr=2 str;t=t1+t2=2 stf+2 str\mathrm{r}=\frac{\mathrm{v}}{\mathrm{t}_2}, \mathrm{t}_2=\frac{\mathrm{v}}{\mathrm{r}}=\frac{2 \mathrm{~s}}{\mathrm{tr}} ; \mathrm{t}=\mathrm{t}_1+\mathrm{t}_2=\frac{2 \mathrm{~s}}{\mathrm{tf}}+\frac{2 \mathrm{~s}}{\mathrm{tr}}
t=(1f+1r)2st⇒t=2s(1f+1r)t=\left(\frac{1}{f}+\frac{1}{r}\right) \frac{2 s}{t} \Rightarrow t=\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}

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