From the adjoining graph, the distance traversed by particle in $4 seconds$ , (in $m$ ) is

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Solution

Distance travelled = Area under $v - t$ graph
= Area I + Area II + Area III + Area IV + Area V
$= (\frac{1}{2}) (O H \times A H) + H G \times A H + \frac{1}{2} (G^{'} C \times B G^{'}) + (G F \times G G^{'}) + (E F \times C F)$
$= (\frac{1}{2}) (1 \times 20) + (1 \times 20) + \frac{1}{2} (1 \times 10) + (1 \times 10) + (1 \times 10)$
$= 10 + 20 + 5 + 10 + 10 = 55 m$

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From the adjoining graph, the distance traversed by particle in 4 seconds, (in m) is

Distance travelled = Area under v−t graph = Area I + Area II + Area III + Area IV + Area V =(12)(OH×AH)+HG×AH+12(G′C×BG′)+(GF×GG′)+(EF×CF) =(12)(1×20)+(1×20)+12(1×10)+(1×10)+(1×10) =10+20+5+10+10=55 m

From the adjoining graph, the distance traversed by particle in $4 seconds$ , (in $m$ ) is