For the velocity-time graph shown in the figu | vedclass

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Question

(A) From the graph:$(A)$ Change in velocity $\Delta v = v(6) - v(0) = 0 - 10 = -10\,m/s$. Thus, $(A \rightarrow r)$.$(B)$ Average acceleration $a_{avg

Vedclass · Accepted Answer

$(A \rightarrow r, B \rightarrow p, C \rightarrow r, D \rightarrow s)$

Vedclass · Answer

(A) From the graph:$(A)$ Change in velocity $\Delta v = v(6) - v(0) = 0 - 10 = -10\,m/s$. Thus, $(A ightarrow r)$.$(B)$ Average acceleration $a_{avg} = \frac{\Delta v}{\Delta t} = \frac{0 - 10}{6 - 0} = -10/6 = -5/3\,m/s^2$. Thus, $(B ightarrow p)$.$(C)$ Total displacement is the area under the $v-t$ graph. From the graph, the line from $t=0$ to $t=4$ is straight. Its slope is $\frac{0 - 10}{2 - 0} = -5\,m/s^2$. At $t=4\,s$, $v = -5 	imes 2 = -10\,m/s$. Area $= 	ext{Area of triangle (0 to 2)} + 	ext{Area of triangle (2 to 6)}$. Area $= (\frac{1}{2} 	imes 2 	imes 10) + (\frac{1}{2} 	imes 4 	imes -10) = 10 - 20 = -10\,m$. Thus, $(C ightarrow r)$.$(D)$ Acceleration at $t=3\,s$ is the slope of the line from $t=0$ to $t=4$, which is $-5\,m/s^2$. Thus, $(D ightarrow s)$.

Column $I$	Column $II$
$(A)$ Change in velocity	$(p)$ $-5/3\,SI \text{ unit}$
$(B)$ Average acceleration	$(q)$ $-20\,SI \text{ unit}$
$(C)$ Total displacement	$(r)$ $-10\,SI \text{ unit}$
$(D)$ Acceleration at $t=3\,s$	$(s)$ $-5\,SI \text{ unit}$

Column $I$	Column $II$
$(A)$ Distance travelled in $3\,s$	$(p)$ $-20$ units
$(B)$ Displacement in $1\,s$	$(q)$ $15$ units
$(C)$ Initial acceleration	$(r)$ $25$ units
$(D)$ Velocity at $4\,s$	$(s)$ $-10$ units

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