(i) Time taken by the man to go to market from his home, $t_1=$ distance $/$ speed $=2.5 / 5=\frac{1}{2} \mathbf{h}$
Time taken by the man to go his home from the market, $t_2=$ distance $/$ speed $=2.5 / 7.5=1 / 3 \mathrm{~h}$
$\therefore$ Total time taken $=t_1+t_2=\frac{1}{2}+1 / 3=5 / 6 \mathrm{~h}=50 \mathrm{~min}$.
(ii) Time interval 0 to $30 \mathrm{~min}$
(a) Average velocity $=$ displacement/time
$=2.5 /(1 / 2)=5 \mathrm{~km} / \mathrm{h}$
(b) Average speed = distance $/$ time $=2.5 /(1 / 2)=5 \mathrm{~km} / \mathrm{h}$
(iii) Time interval 0 to $50 \mathrm{~min}$
Total distance travelled $=2.5+2.5=5 \mathrm{~km}$
Total displacement $=$ zero
(a) Average velocity
$=$ displacement $/$ time $=0$
(b) Average speed $=$ distance $/$ time
$=5 /(5 / 6)=6 \mathrm{~km} / \mathrm{h}$
(iv) Time interval 0 to $40 \mathrm{~min}$
Distance moved in $30 \mathrm{~min}$
(from home to market) $=2.5 \mathrm{~km}$
Distance moved in $10 \mathrm{~min}$ (from market to home) with
a speed $7.5 \mathrm{~km} / \mathrm{h}=7.5 \times 10 / 60=1.25 \mathrm{~km}$
So displacement $=2.5-1.25=1.25 \mathrm{~km}$
Distance travelled $=2.5+1.25=3.75 \mathrm{~km}$
(a) Average velocity $=1.25 /(40 / 60)$
$=1.875 \mathrm{~km} / \mathrm{h}$
(b) Average speed $=3.75 /(40 / 60)$
$=5.625 \mathrm{~km} / \mathrm{h}$