Q. For motion in a plane with constant acceleration $\overrightarrow{a},$ initial velocity ${\overrightarrow{v}}_{0}$ and final velocity $\overrightarrow{v}$ after time $t$, we have
A) $\overrightarrow{v}.\left( \overrightarrow{v} - \overrightarrow{a}t \right) = {\overrightarrow{v}}_{0}.({\overrightarrow{v}}_{0} + \overrightarrow{a}t)$
B) $\overrightarrow{v}.{\overrightarrow{v}}_{0} = at^{2}$
C) $\overrightarrow{v}.{\overrightarrow{v}}_{0} = \overrightarrow{v}.{\overrightarrow{v}}_{0}t$
D) ${\overrightarrow{v}}_{0}.{\overrightarrow{v}}_{0} = \overrightarrow{a}.{\overrightarrow{v}}_{0}t$
Explanation
(a) Since $v^{2} - v_{0}^{2} = 2\overrightarrow{a}.\overrightarrow{s} = 2\overrightarrow{a}.\left( \frac{\overrightarrow{v} + {\overrightarrow{v}}_{0}}{2} \right)t$
or $\overrightarrow{v}.\overrightarrow{v} - {\overrightarrow{v}}_{0}.{\overrightarrow{v}}_{0} = \left( \overrightarrow{v} + {\overrightarrow{v}}_{0} \right).\overrightarrow{a}t$
or $\overrightarrow{v}.\left( \overrightarrow{v} - \overrightarrow{a\ }t \right) = {\overrightarrow{v}}_{0}.({\overrightarrow{v}}_{0} + \overrightarrow{a}t)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.