A particle starts from rest and moves in a straight line. It travels a distance 2 L with uniform acceleration and then moves with a constant velocity a further distance of L. Finally, it comes to rest after moving a distance of 3L under uniform retardation. Then the ratio of average speed to the…

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A particle starts from rest and moves in a straight line. It travels a distance 2 L with uniform acceleration and then moves with a constant velocity a further distance of L. Finally, it comes to rest after moving a distance of 3L under uniform retardation. Then the ratio of average speed to the maximum speed $\left(\frac{\bar{V}}{V_m}\right)$ of the particle is

Aicharya · Accepted Answer

The $v$-t graph of the particle is

$\begin{aligned} & 2 L=\frac{1}{2} v_{\max } \times t_1 \Rightarrow t_1=\frac{4 L}{v_{\max }} \\ & L=v_{\max } \times t_2 \Rightarrow t_2=\frac{L}{v_{\max }}\end{aligned}$ $\begin{aligned} & 3 \mathrm{~L}=\frac{1}{2} \times \mathrm{v}_{\max } \times \mathrm{t}_3 \Rightarrow \mathrm{t}_3=\frac{6 \mathrm{~L}}{\mathrm{v}_{\max }} \\ & \therefore \quad \text { Average speed, } \overline{\mathrm{v}}=\frac{\text { Total distance }}{\text { Total time }} \\ & \therefore \quad \overline{\mathrm{v}}=\frac{2 \mathrm{~L}+\mathrm{L}+3 \mathrm{~L}}{\mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3}=\left(\frac{6 \mathrm{~L}}{11 \mathrm{~L}}\right) \mathrm{v}_{\max } \\ & \therefore \frac{\overline{\mathrm{v}}}{\mathrm{v}_{\max }}=\frac{6}{11}\end{aligned}$

A particle starts from rest and moves in a straight line. It travels a distance 2 L with uniform…

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