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2.Motion in Straight Line
medium
For the velocity-time graph shown in the figure, in a time interval from $t=0$ to $t=6\,s$, match the following columns.
| Colum $I$ | Colum $II$ |
| $(A)$ Change in velocity | $(p)$ $-5 / 3\,Sl$ unit |
| $(B)$ Average acceleration | $(q)$ $-20\,SI$ unit |
| $(C)$ Total displacement | $(r)$ $-10\,SI$ unit |
| $(D)$ Acceleration at $t=3\,s$ | $(s)$ $-5\,SI$ unit |
A
$(A \rightarrow r, B \rightarrow p, C \rightarrow r, D \rightarrow s)$
B
$(A \rightarrow p, B \rightarrow r, C \rightarrow r, D \rightarrow s)$
C
$(A \rightarrow r, B \rightarrow r, C \rightarrow p, D \rightarrow s)$
D
$(A \rightarrow p, B \rightarrow p, C \rightarrow r, D \rightarrow s)$
Solution
(a)
$v_i=+10\,m / s \text { and } v_f=0$
$\therefore \quad \Delta v=v_f-v_i=-10\,m / s$
$a_{2 v} =\frac{\Delta u}{\Delta t}=\frac{-10}{6}$
$=\frac{-5}{3}\,m / s ^2$
Total displacement $=$ area under $v-t$ graph (with sign)
And acceleration $=$ slope of $v-t$ graph
Standard 11
Physics
