\(where\)
\(\bar{v} = avg\,\,velocity\)
\(\Delta\,x = dp\,\,(displacement) = change\,\,in\,\, position\,\, and\,\, direction\)
\(so\)
\(\Delta\,\,x = dp = x_{f} - x_{o} = \,\,x\,(final) - x\,(initial)\)
\(and\)
\(\Delta\,t = change\,\,in\,\,time\)
\(so\)
\(\Delta\,\,t = t_{f} - t_{o} = \,\,t\,(final) - t\,(initial) \)
\(m = meter,\,\, km = kilometer,\,\, s = second,\,\, min = minute,\,\,N = north\)
\(\bar{v} = \dfrac{\Delta\, x}{\Delta\, t}\)A car travels 90 degrees north, going \(1000\, m\, (1\, km)\,\) in \(60\, s\, (1\, min)\). What is the avg velocity?
\(and\)
\(\Delta\,\,x = dp = x_{f} - x_{o} = 1000\,\,m\,\,90^{\circ}\,\,N - 0\,\,m = 1000\,\,m\,\,90^{\circ}\,\,N\)
\(\Delta\,\,t = t_{f} - t_{o} = 60\,\,s - 0\,\,s = 60\,\,s\)
\(so\)
\(\bar{v} = \dfrac{1000 \,\,m\,\,90^{\circ}\,\,N}{60\, s} = \dfrac{16.7\,\,m\,\,90^{\circ}\,\, N}{s}\)
