A motorist drives south at 20.0 m/sfor 3.00 min, then turns west and travels at

Question

A motorist drives south at 20.0 m/sfor 3.00 min, then turns west and travels at 25.0 m/s for2.00 min, and finally travelsnorthwest at 30.0 m/s for 1.00 min. For this 6.00 min trip, find the following values. (a) total vector displacement
                          m (magnitude) ° south of west
(b) average speed

                       m/s
(c) average velocity

m/s (magnitude) 5° south of west (a) total vector displacement
                          m (magnitude) ° south of west
(b) average speed

                       m/s
(c) average velocity

m/s (magnitude) 5° south of west

Explanation / Answer

Therefore x1 = 0m and y1 = -3600m
Next displacement is (25m/s)(120s)toward west Therefore x2 = -3000m and y2 =0m
Next displacement is (30m/s)(60s)toward northwest Therefore x3 = -(1800m)(cos45) =-1272.79m and y3 = 1800m)(sin45) =1272.79m
Then X-component of total displacement is 0m + (-3000m) +(-1272.79m) = -4272.79m
And Y-component of total displacement is (-3600)m + 0m +(1272.79m) = -2327.21m
Then the magnitude of the total displacement issqrt[(-4272.79m)2 + (-2327.21m)2] =4865.45m
direction is tan-1(2327.21/4272.79)South of west =28.57o south of west
Total distance traveled is (20m/s)(180s)+ (25m/s)(120s) + (30m/s)(60s) = 8400m total time of travel t = 180s + 120s + 60s = 360s Then the average speed (8400m)/(360s) = 23.33m/s

Average velocity = (4865.45m)/(360s)       = 13.5m/sat 28.57o south of west

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