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(A) Let the total distance be $x$. The first half distance is $x/2$ covered with speed $v_1 = 3 \, m/s$. The time taken is $t_1 = \frac{x/2}{3} = \fra

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(A) Let the total distance be $x$. The first half distance is $x/2$ covered with speed $v_1 = 3 \, m/s$. The time taken is $t_1 = \frac{x/2}{3} = \frac{x}{6}$.The second half distance $x/2$ is covered in two equal time intervals,say $t_2$ each. Let the speeds be $v_2 = 4.5 \, m/s$ and $v_3 = 7.5 \, m/s$.Distance covered in the second half is $(v_2 \cdot t_2) + (v_3 \cdot t_2) = x/2$.$(4.5 + 7.5) t_2 = x/2 \Rightarrow 12 t_2 = x/2 \Rightarrow t_2 = x/24$.The total time taken for the second half is $2 t_2 = 2(x/24) = x/12$.Total time $T = t_1 + 2 t_2 = \frac{x}{6} + \frac{x}{12} = \frac{2x + x}{12} = \frac{3x}{12} = \frac{x}{4}$.Average speed $v_{avg} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{x}{x/4} = 4 \, m/s$.

$A$ particle moving in a straight line covers half the distance with a speed of $3 \, m/s$. The other half of the distance is covered in two equal time intervals with speeds of $4.5 \, m/s$ and $7.5 \, m/s$ respectively. The average speed of the particle during this motion is...... $\, m/s$

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