Question

For the velocity-time graph shown in figure, in a time interval from $t=0$ to $t=6\text{s}$, match the following

$\begin{array}{cc}\text{Table-1}& \text{Table-2}\\ \text{(A) Change in velocity}& \text{(P)}-\frac{5}{3}\text{SI unit}\\ \text{(B) Average acceleration}& \text{(Q)}-20\text{SI unit}\\ \text{(C) Total displacement}& \text{(R)}-10\text{SI unit}\\ \text{(D) Acceleration at}t=3\text{s}& \text{(S)}-5\text{SI unit}\end{array}$

• A. $\left(A\right)\to \left(S\right);\left(B\right)\to \left(P\right);\left(C\right)\to \left(R\right);\left(D\right)\to \left(Q\right)$
• B. $\left(A\right)\to \left(S\right);\left(B\right)\to \left(P\right);\left(C\right)\to \left(S\right);\left(D\right)\to \left(R\right)$
• C. $\left(A\right)\to \left(R\right);\left(B\right)\to \left(P\right);\left(C\right)\to \left(R\right);\left(D\right)\to \left(S\right)$
• D. $\left(A\right)\to \left(R\right);\left(B\right)\to \left(P\right);\left(C\right)\to \left(S\right);\left(D\right)\to \left(P\right)$

Answer 1

The correct option is C $\left(A\right)\to \left(R\right);\left(B\right)\to \left(P\right);\left(C\right)\to \left(R\right);\left(D\right)\to \left(S\right)$

Initial velocity $v_i=+10\text{m/s}$
and final velocity $v_f=0$
$\therefore$ Change in velocity $=\Delta v=v_f-v_i=-10\text{m/s}$
$\text{Average acceleration}=\frac{\text{Total change in velocity}}{\text{Total time}}$
$a_{av}=\frac{\Delta v}{\Delta t}=\frac{-10}{6}=\frac{-5}{3}{\text{m/s}}^2$
Total displacement $=$ area under $v-t$ graph (with sign)
Area under the $v-t$ graph from $t=0$ to $t=6\text{s}$ is given by
$A=\frac{1}{2}\times 10\times 2-\frac{1}{2}\times 4\times 10=-10\text{m}$
Acceleration $=$ slope of $v-t$ graph.
$\Rightarrow$ Slope of the graph at $(t=3\text{s})=\frac{-10-0}{4-2}=-5{\text{m/s}}^2$