Question

A particle is moving such that its position coordinate (x,y) are (2m,3m) at time t=0,(6m,7m) at time t=2s and (13m,14m) at time t=5s. Average velocity vector from t=0 to t=5s

Answer 1

Displacement=(11m,11m) at
Time=5s
So average velocity vector =(2.2,2.2)m/s
Or in vector form- 2.2icab+2.2jcab

Answer 2

Answer: Average velocity vector from [t=0 to 5] $=\quad \frac { 11 }{ 5 } \overrightarrow { i } +\frac { 11 }{ 5 } \overrightarrow { j }$

             ${ v }_{ average }\quad =\quad \frac { dx }{ dt }$

             $dx\quad =\quad Final\quad position\quad -\quad Initial\quad position$

             $dx\quad =\quad \left( 13\overrightarrow { i } +14\overrightarrow { j } \right) \quad -\quad \left( 2\overrightarrow { i } +3\overrightarrow { j } \right)$

             $dx\quad =\quad \left( 13\quad -\quad 2 \right) \overrightarrow { i } \quad +\quad \left( 14\quad -\quad 3 \right) \overrightarrow { j } \\ dx\quad =\quad 11\overrightarrow { i } +11\overrightarrow { j }$

              $dt\quad =\quad 5\quad sec\quad -\quad 0\quad sec\quad =\quad 5\quad sec\\ { v }_{ average }\quad =\quad \frac {11\overrightarrow{ i }+11\overrightarrow{ j }}{ 5 }$

               $Average\quad velocity\quad vector\quad =\quad \frac { 11 }{ 5 }\overrightarrow { i } +\frac { 11 }{ 5 } \overrightarrow { j }$