A body travels a distance $s$ in $t$ seconds. It starts from rest and ends at rest. In the first part of the…

A body travels a distance $s$ in $t$ seconds. It starts from rest and ends at rest. In the first part of the journey, it moves with constant acceleration $\mathrm{f}$ and in the second part with constant retardation $\mathrm{r}$. The value of $\mathrm{t}$ is given by
  1. $\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}$
  2. $2 s\left(\frac{1}{f}+\frac{1}{r}\right)$
  3. $\frac{2 \mathrm{~s}}{\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{r}}}$
  4. $\sqrt{2 s(f+r)}$

Solution

Portion OA, OB corresponds to motion with acceleration ' $\mathrm{f}$ ' and retardation ' $r$ ' respectively. Area of $\triangle O A B=S$ and $O B=t$. Let $O L=t_1$, $\mathrm{LB}=\mathrm{t}_2$ and $\mathrm{AL}=\mathrm{v}, \mathrm{S}=\frac{1}{2} \mathrm{OB} \cdot \mathrm{AL}=\frac{1}{2} \mathrm{t} \cdot \mathrm{v} ; \mathrm{v}=\frac{2 \mathrm{~S}}{\mathrm{t}}$ Also, $\mathrm{f}=\frac{\mathrm{v}}{\mathrm{t}_1}, \mathrm{t}_1=\frac{\mathrm{v}}{\mathrm{f}}=\frac{2 \mathrm{~s}}{\mathrm{tf}}$ and $\mathrm{r}=\frac{\mathrm{v}}{\mathrm{t}_2}, \mathrm{t}_2=\frac{\mathrm{v}}{\mathrm{r}}=\frac{2 \mathrm{~s}}{\mathrm{tr}} ; \mathrm{t}=\mathrm{t}_1+\mathrm{t}_2=\frac{2 \mathrm{~s}}{\mathrm{tf}}+\frac{2 \mathrm{~s}}{\mathrm{tr}}$ $t=\left(\frac{1}{f}+\frac{1}{r}\right) \frac{2 s}{t} \Rightarrow t=\sqrt{2 s\left(\frac{1}{f}+\frac{1}{r}\right)}$

Asked in: JEE Main 2003

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