
The area under the velocity time graph gives the displacement. The area for t=0stot=25sis
(Area)=(21×5×10)+(5×10)+(21×30×5)+(21×20×5)−(21×20×5)
⇒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 =(21×5×10)+(5×10)+(21×30×5)+(21×20×5)+(21×20×5)=(200+50)=250m
Therefore, required ratio
displacement distance =150250=(35)