Correct Option is (D)

Area under the graph from $t=0$ to $t=20\sec =200\mathrm{m}$

Area under the graph from $\mathrm{t}=20$ to $t=25\sec =50\mathrm{m}$

So distance covered $=(200+50)\mathrm{m}=250\mathrm{m}$

Displacement $=(200-50)\mathrm{m}=150\mathrm{m}$

$\therefore$ $\frac{\text{ distance }}{\text{ displacement }}=\frac{250}{150}=\left(\frac{5}{3}\right)$

Correct Option is (B)

From the given graphs, it can be seen

$\begin{array}{c}{\text{ slope }}_{\mathrm{B}}>{\text{ slope }}_{\mathrm{A}}\\ \\ v_{\mathrm{B}}>v_{\mathrm{A}}\end{array}$

$\therefore$ Statement (E) is correct.

Also, $x_{\mathrm{A}}<x_{\mathrm{B}}$, A lives closer to school.

$\therefore$ Statement $(\mathrm{A})$ is correct.

Correct Option is (C)

Area under velocity time graph gives displacement of body in given time.

Area under acceleration time graph gives change in velocity in the given time.

So Statement I false but Statement II True

Correct Option is (B)

$\begin{array}{rl}& \frac{\mathrm{dx}}{\mathrm{dt}}=\text{ slope }\ge 0\text{ always increasing }\\ \\ & (\mathrm{A}\text{ - II) }\\ \\ & \frac{\mathrm{dx}}{\mathrm{dt}}<0\text{; and at }\mathrm{t}\to \mathrm{\infty }\frac{\mathrm{dx}}{\mathrm{dt}}\to 0\\ \\ & \text{ (B - IV) }\\ \\ & \frac{\mathrm{dx}}{\mathrm{dt}}>0\text{ for first half }\frac{\mathrm{dx}}{\mathrm{dt}}<0\text{ for second half. }\\ \\ & (\mathrm{C}\text{ - III) }\\ \\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\text{ constant }\\ \\ & \text{ (D - I) }\end{array}$

Correct Option is (B)

From $v$-t graph

Displacement $=$ Area under curve considering sign also.

Distance $=$ Area under curve considering only magnitude

Displacement $=\mathrm{\Sigma }$area $=8$$\times$$2-4$$\times$$2+4$$\times$$4-2$$\times$$4=16\mathrm{m}$

Distance $=\mathrm{\Sigma }\mid$ area $\mid =48\mathrm{m}$

Displacement : distance $=1:3$