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(A) The $v$-$t$ graph of the particle is shown in the figure.$1$. For the first part (uniform acceleration):Distance $2L = \frac{1}{2} 	imes V_m 	im

Vedclass · Accepted Answer

$\frac{6}{11}$

Vedclass · Answer

(A) The $v$-$t$ graph of the particle is shown in the figure.$1$. For the first part (uniform acceleration):Distance $2L = \frac{1}{2} 	imes V_m 	imes t_1 \implies t_1 = \frac{4L}{V_m}$$2$. For the second part (constant velocity):Distance $L = V_m 	imes t_2 \implies t_2 = \frac{L}{V_m}$$3$. For the third part (uniform retardation):Distance $3L = \frac{1}{2} 	imes V_m 	imes t_3 \implies t_3 = \frac{6L}{V_m}$$4$. Average speed $\bar{V} = \frac{	ext{Total distance}}{	ext{Total time}}$Total distance $= 2L + L + 3L = 6L$Total time $= t_1 + t_2 + t_3 = \frac{4L}{V_m} + \frac{L}{V_m} + \frac{6L}{V_m} = \frac{11L}{V_m}$$5$. Therefore,$\bar{V} = \frac{6L}{11L/V_m} = \frac{6}{11} V_m$$\frac{\bar{V}}{V_m} = \frac{6}{11}$

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