The velocity-time graph of a body moving in a straight line is...

Area of upper rectangles, which will give us magnitude of the area under the velocity-time graph,

${A}_{1} = 8 \times 2 + 4 \times 4 = 32$

Area of lower rectangles, which will give us magnitude of the area under the velocity-time graph,

${A}_{2} = 4 \times 2 + 4 \times 2 = 16$

Therefore,

Displacement = {A}_{1} - {A}_{2} = 32 - 16 = 16$<br/>Distance = {A}_{1} + {A}_{2} = 32 + 16 = 48$

Hence, the required ratio would be

$= \dfrac{16}{48} = \dfrac{1}{3}$