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(B) Let the total time be $T$. The first half time is $t = T/2$.For particle $A$:Velocity at $t = T/2$ is $v_A = 0 + 2(T/2) = T$.Distance in first hal

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$B$

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(B) Let the total time be $T$. The first half time is $t = T/2$.For particle $A$:Velocity at $t = T/2$ is $v_A = 0 + 2(T/2) = T$.Distance in first half $s_{A1} = 0 + 1/2(2)(T/2)^2 = T^2/4$.Distance in second half $s_{A2} = v_A(T/2) + 1/2(4)(T/2)^2 = T(T/2) + 2(T^2/4) = T^2/2 + T^2/2 = T^2$.Total distance $S_A = T^2/4 + T^2 = 5T^2/4$.For particle $B$:Velocity at $t = T/2$ is $v_B = 0 + 4(T/2) = 2T$.Distance in first half $s_{B1} = 0 + 1/2(4)(T/2)^2 = T^2/2$.Distance in second half $s_{B2} = v_B(T/2) + 1/2(2)(T/2)^2 = 2T(T/2) + 1(T^2/4) = T^2 + T^2/4 = 5T^2/4$.Wait,let's re-evaluate the graph. The area under the $v-t$ graph represents displacement. Since both particles start from rest and reach the same final velocity at time $T$ $(v_{final} = a_1(T/2) + a_2(T/2) = (2+4)(T/2) = 3T)$,the area under the curve for $B$ is clearly larger because it gains velocity faster in the first half.Thus,particle $B$ covers a larger distance.

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