Circle-velocity

$Since,\quad the\quad body\quad is\quad initially\quad at\quad rest,\quad \\ So,\quad it's\quad intial\quad angular\quad velocity\quad (\omega )\quad =\quad 0\\ Now,\quad its\quad angular\quad acceleration\quad (\alpha )\quad =\quad \frac { \pi }{ 4 } \\ Since,\quad it\quad covers\quad quarter\quad a\quad circle,\quad angular\\ displacement\quad (\theta )\quad =\quad \frac { \pi }{ 2 } \\ \\ Using\quad the\quad first\quad equation\quad of\quad motion\quad of\quad circular\quad dynamics,\\ \theta \quad =\quad \omega t\quad +\quad \frac { 1 }{ 2 } \alpha { t }^{ 2 },\quad where\quad t\quad is\quad in\quad seconds\\ \frac { \pi }{ 2 } \quad =\quad 0\quad +\quad \frac { 1 }{ 2 } \frac { \pi }{ 4 } { t }^{ 2 }\\ Cancelling\quad like\quad terms,\quad \\ t\quad =\quad 2\quad seconds.\\ \\ Now,\quad let\quad the\quad inital\quad position\quad of\quad the\quad body\quad be\quad 'A'.\\ Let,\quad the\quad center\quad of\quad the\quad circle\quad be\quad 'O'\quad and\quad the\quad final\\ position\quad be\quad 'B'.\\ In\quad \triangle OAB,\quad \angle AOB\quad =\quad \frac { \pi }{ 2 } \\ Since\quad it\quad is\quad an\quad isosceles\quad triangle,\quad the\quad other\quad two\quad angles\\ are\quad \frac { \pi }{ 4 } rad.\quad each.\\ So,\quad using\quad sine\quad law,\\ \frac { \bar { OA } }{ \sin { \frac { \pi }{ 4 } } } =\frac { \bar { AB } }{ \sin { \frac { \pi }{ 2 } } } \\ Now,\quad \bar { OA } \quad is\quad the\quad radius\quad of\quad the\quad circle\quad =\quad \sqrt { 2 } m.\\ Using\quad the\quad values,\\ \frac { \sqrt { 2 } }{ \frac { 1 }{ \sqrt { 2 } } } =\frac { \bar { AB } }{ 1 } \\ \bar { AB } \quad =\quad 2m.\\ Now,\quad \bar { AB } \quad is\quad the\quad linear\quad displacement\quad of\quad the\quad particle\\ from\quad 'A'\quad to\quad 'B'.\\ Average\quad velocity\quad =\quad \frac { Total\quad linear\quad displacement }{ time\quad taken } \\ { v }_{ avg. }=\quad \frac { 2 }{ 2 } m/s\\ { v }_{ avg. }=\quad 1m/s$