The velocity versus time graph of a body movi | vedclass

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Question

(D) The distance covered by the body is equal to the area under the velocity-time $(v-t)$ graph.For the interval $t = 0 \,s$ to $t = 3 \,s$, the area

Vedclass · Accepted Answer

The distance moved by the body during $0$ to $3 \,s$ is $6 \,m$.

Vedclass · Answer

(D) The distance covered by the body is equal to the area under the velocity-time $(v-t)$ graph.For the interval $t = 0 \,s$ to $t = 3 \,s$, the area is a triangle with base $b = 3 \,s$ and height $h = 4 \,m/s$.Distance $= 	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 3 	imes 4 = 6 \,m$.Thus, option $(D)$ is correct.For the interval $t = 0 \,s$ to $t = 2 \,s$, the distance is $\frac{1}{2} 	imes 2 	imes 4 = 4 \,m$.The acceleration in the interval $t = 0 \,s$ to $t = 2 \,s$ is the slope of the graph: $a = \frac{4-0}{2-0} = 2 \,ms^{-2}$.The acceleration in the interval $t = 2 \,s$ to $t = 3 \,s$ is the slope of the graph: $a = \frac{0-4}{3-2} = -4 \,ms^{-2}$.

Column $I$	Column $II$
$(A)$ $M$	$(p)$ $A^{-1}$
$(B)$ $N$	$(q)$ $R^{-1}$
$(C)$ $P$	$(r)$ $A$
$(D)$ $Q$	$(s)$ $R$

The velocity versus time graph of a body moving in a straight line is as shown in the figure below.

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