Form the v - t graph shown, the ratio of distance to displacement...

The area under the velocity time graph gives the displacement. The area for $t = 0s\text{to}t = 25s$is

$( Area ) = \left(\dfrac{1}{2} \times 5 \times 10\right) + \left(5 \times 10\right) + \left(\dfrac{1}{2} \times 30 \times 5\right) + \left(\dfrac{1}{2} \times 20 \times 5\right) - \left(\dfrac{1}{2} \times 20 \times 5\right)$

$⇒ Area = 25 + 50 + 75 + 50 - 50 = 150m$
Hence, the displacement $= 150m$

The distance is defined as the path taken by the object to travel from initial to final position. Also note that displacement is a vector quantity, but distance is a scalar quantity.
Distance $= \left(\dfrac{1}{2} \times 5 \times 10\right) + \left(5 \times 10\right) + \left(\dfrac{1}{2} \times 30 \times 5\right) + \left(\dfrac{1}{2} \times 20 \times 5\right) + \left(\dfrac{1}{2} \times 20 \times 5\right) = ( 200 + 50 ) = 250m$

Therefore, required ratio
$\dfrac{\text{distance}}{\text{displacement}} = \dfrac{250}{150} = \left(\dfrac{5}{3}\right)$