What is the total distance covered by the body at the end of the journ

1.	What is the total distance covered by the body at the end of the journey as shown in the graph.75 m67.5 m100 mnone of the above
Answer» Given: A velocity time graph is given . To Find: The total displacement covered by the body at the end of the journey. Concept Used: As we know that the area of ( v - t ) graph gives US displacement , so we will be finding the area of the graph . Answer: (For graph refer to the attachment:) We can here find the area of the graph in THREE steps , Firstly finding area of ∆ ABE . Secondly finding area of rectangle BCEF. Thirdly finding area of ∆ DFC . Note:On finding area power of dimension will be 1 . $\red{\bf{\leadsto Step\:1}}$ Finding area of ∆ ABE Here , HEIGHT = 15m/s. Base = 2s $\sf{\implies ar(\Delta ABE )= \dfrac{1}{2}\times (base)\times (height) }$ $\sf{\implies ar(\Delta ABE )= \dfrac{1}{2}\times 2 s\times 15m/s}$ $\boxed{\red{\bf{\longmapsto ar(\triangle ABE )= 15m}}}$ $\rule{200}5$ $\red{\bf{\leadsto Step\:2}}$ Finding area of rectangle BCEF Here , BREADTH = (5-2)s = 3s. Length = 15m/s. $\sf{\implies ar(BCEF) = (length)\times (breadth)}$ $\sf{\implies ar(BCEF) = 3s \times 15m/s }$ $\boxed{\red{\bf{\longmapsto ar( BCEF)= <klux>45M</klux>}}}$ $\rule{200}5$ $\red{\bf{\leadsto Step\:3}}$ Finding area of ∆ DFC. Here , Height = 15m/s. Base = (6-5)s = 1s. $\sf{\implies ar(\Delta DFC) = \dfrac{1}{2}\times (base)\times (height) }$ $\sf{\implies ar(\Delta DFC)= \dfrac{1}{2}\times 1 s\times 15m/s}$ $\boxed{\red{\bf{\longmapsto ar(\triangle DFC )= 7.5m}}}$ $\rule{200}5$ Hence total area = displacement = ar(∆DFC) + ar(BCEF) + ar(∆ABE). = 15m + 45m + 7.5m = 67.5m. Hence the total displacement is 67.5m .