Question

Study the following $v–t$ graphs in $\text{Column I}$ carefully and match appropriately with the statements given in $\text{Column II}$. Assume that motion takes place from time $0$ to $T$.

• A. $\text{i}\to \left(q\right),\left(s\right);\text{ii}\to \left(p\right),\left(s\right);\text{iii}\to \left(r\right);\text{iv}\to \left(p\right)$
• B. $\text{i}\to \left(q\right),\left(s\right);\text{ii}\to \left(p\right);\text{iii}\to \left(r\right),\left(s\right);\text{iv}\to \left(p\right),\left(s\right)$
• C. $\text{i}\to \left(p\right),\left(s\right);\text{ii}\to \left(q\right),\left(s\right);\text{iii}\to \left(r\right);\text{iv}\to \left(p\right),\left(q\right)$
• D. $\text{i}\to \left(p\right),\left(s\right);\text{ii}\to \left(q\right);\text{iii}\to \left(r\right),\left(s\right);\text{iv}\to \left(q\right)$

Answer 1

The correct option is A $\text{i}\to \left(q\right),\left(s\right);\text{ii}\to \left(p\right),\left(s\right);\text{iii}\to \left(r\right);\text{iv}\to \left(p\right)$

i. Area of $v–t$ graph lies below time axis, so displacement is negative. But slope is positive, so acceleration is positive.
ii. Area of $v-t$ graph lies above time axis, so displacement is positive. And slope is positive, so acceleration is also positive.
iii. Displacement is zero, because half area is above time axis and half below. Slope is negative, so acceleration is negative.
iv. Area of $v-t$ graph lies above time axis, so displacement is positive. And slope is negative, so acceleration is negative.