# from the graph find the slop at 5 sec and area enclosed by the graph at sec

The slop at 5 sec

Value of velocity at 5 sec is 50 m/sec. Therefore slop = 50/5 = 10m/sec2

The value of valocity at t = 4 sec = 40m/sec. Therefore the area enclosed

$=\frac{1}{2}bh=\frac{1}{2}\times4\times40=80m$

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