A motorist drives south at 22.0 m/s for 3.00 min, then turnswest and travels at

Question

A motorist drives south at 22.0 m/s for 3.00 min, then turnswest and travels at 25.0 m/s for 2.00 min, and finally travelsnorthwest at 30.0 m/s for 1.00 min. For this 6.00 min trip, findthe following values. My degrees for part (c) keeps coming out wrong. (a) total vector displacement

         5046.68 m(magnitude)  
          32.16° south of west (b) average speed          24.33 m/s (c) average velocity         14.02 m/s (magnitude)                   °south of west A motorist drives south at 22.0 m/s for 3.00 min, then turnswest and travels at 25.0 m/s for 2.00 min, and finally travelsnorthwest at 30.0 m/s for 1.00 min. For this 6.00 min trip, findthe following values. My degrees for part (c) keeps coming out wrong. (a) total vector displacement

         5046.68 m(magnitude)  
          32.16° south of west (b) average speed          24.33 m/s (c) average velocity         14.02 m/s (magnitude)                   °south of west                   °south of west

Explanation / Answer

1 st case : --------- speed v = 22.0 m/s time t= 3.00 min= 3*60 s = 180 s displacement   S = v t                          = 3960 m displacement vector S = 3960 *(-j ) Since it is in south direction case2 : ------- velocity v ' = 25.0 m/s time t ' = 2.00 min = 2 * 60 s = 120 s displacement S ' = v ' t '                           = 3000 m displacement vector S ' = 3000* (-i ) since it is in west direction Here i , j are unit vectors in east and northdirections respectively Case 3: ------- velocity v " = 30.0 m/s time t" = 1.00 min= 60 s displacement S " = v " t "                          = 1800 m displacement vector S " = 1800 cos 45 * (-i ) +1800 sin 45 ( j )                                     = 1272.79 ( -i ) + 1272.79 j Total time T = 6.00 min= 6 * 60 = 360 s (a). Total vector displacement= S + S ' +S "                                              = -4272.79 i +-2687.21 j magnitude = sqrt of [-4272.79^2+(-2687.21)^2]                 = 5047.55 m

Let the displacement vector makes an angle with west thentan = 2687.21 / 4272.79                                                     = 32.16 degrees (b) average speed = total distance / total time total distance = 3960 m +3000 m + [ -1272.79^2 +1272.79^2 ]                    = [3960m + 3000 m + 1800 m] = 8760 m So, average speed = = 8760 m / 360 s= 24.33 m/ s (c).Average velocity = total displacement / total time                               = {[ -4272.79^2 +-2687.21^2]} / 360 s                                =14 m / s         speed v = 22.0 m/s time t= 3.00 min= 3*60 s = 180 s displacement   S = v t                          = 3960 m displacement vector S = 3960 *(-j ) Since it is in south direction case2 : ------- velocity v ' = 25.0 m/s time t ' = 2.00 min = 2 * 60 s = 120 s displacement S ' = v ' t '                           = 3000 m displacement vector S ' = 3000* (-i ) since it is in west direction Here i , j are unit vectors in east and northdirections respectively Case 3: ------- velocity v " = 30.0 m/s time t" = 1.00 min= 60 s displacement S " = v " t "                          = 1800 m displacement vector S " = 1800 cos 45 * (-i ) +1800 sin 45 ( j )                                     = 1272.79 ( -i ) + 1272.79 j Total time T = 6.00 min= 6 * 60 = 360 s (a). Total vector displacement= S + S ' +S "                                              = -4272.79 i +-2687.21 j magnitude = sqrt of [-4272.79^2+(-2687.21)^2]                 = 5047.55 m

Let the displacement vector makes an angle with west thentan = 2687.21 / 4272.79                                                     = 32.16 degrees (b) average speed = total distance / total time total distance = 3960 m +3000 m + [ -1272.79^2 +1272.79^2 ]                    = [3960m + 3000 m + 1800 m] = 8760 m So, average speed = = 8760 m / 360 s= 24.33 m/ s (c).Average velocity = total displacement / total time                               = {[ -4272.79^2 +-2687.21^2]} / 360 s                                =14 m / s        

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