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(C) From the graph,the initial velocity $u = 1\, m/s$ at $t = 0\, s$ and final velocity $v = 2\, m/s$ at $t = 1\, s$.The acceleration $a$ is the slope

Vedclass · Accepted Answer

average speed during $1^{st}\, second$ is $1.5\, m/s$

Vedclass · Answer

(C) From the graph,the initial velocity $u = 1\, m/s$ at $t = 0\, s$ and final velocity $v = 2\, m/s$ at $t = 1\, s$.The acceleration $a$ is the slope of the $v-t$ graph:$a = 	an(45^{\circ}) = 1\, m/s^2$.The distance covered is the area under the $v-t$ graph,which is a trapezoid:$s = 	ext{Area} = \frac{1}{2} 	imes (u + v) 	imes t = \frac{1}{2} 	imes (1 + 2) 	imes 1 = 1.5\, m$.The average speed is given by:$v_{avg} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{1.5\, m}{1\, s} = 1.5\, m/s$.Thus,option $C$ is correct.

From the $v-t$ graph,the

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