A particle starts from rest and traverses a d | vedclass

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Question

(C) Let $v_m$ be the maximum speed attained by the particle.$1$. For the first part (acceleration): $v_m^2 = 0 + 2a_1 l \implies a_1 = v_m^2 / (2l)$.

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$3$

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(C) Let $v_m$ be the maximum speed attained by the particle.$1$. For the first part (acceleration): $v_m^2 = 0 + 2a_1 l \implies a_1 = v_m^2 / (2l)$. The time taken is $t_1 = v_m / a_1 = 2l / v_m$.$2$. For the second part (uniform motion): The distance is $2l$ at speed $v_m$. The time taken is $t_2 = 2l / v_m$.$3$. For the third part (retardation): $0 = v_m^2 - 2a_2(3l) \implies a_2 = v_m^2 / (6l)$. The time taken is $t_3 = v_m / a_2 = 6l / v_m$.Total distance $D = l + 2l + 3l = 6l$.Total time $T = t_1 + t_2 + t_3 = (2l / v_m) + (2l / v_m) + (6l / v_m) = 10l / v_m$.Average speed $v_{av} = D / T = 6l / (10l / v_m) = 0.6 v_m = (3/5) v_m$.Therefore,the ratio $v_{av} / v_m = 3/5$.

Column $I$	Column $II$
$(A)$ Constant positive acceleration	$(p)$ Speed may increase
$(B)$ Constant negative acceleration	$(q)$ Speed may decrease
$(C)$ Constant displacement	$(r)$ Speed is zero
$(D)$ Constant slope of $a-t$ graph	$(s)$ Speed must increase

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