University of Ghana
posted by Daniel .
• A car moves 26 km due east then 14 km due north. It then turns along a path running northwest with 15 km, then 17 km due west. If the time for the entire journey is 2 hrs.
• Find;
• (a) the car’s average speed in m/s
• (b) the car’s average velocity in km/h

Average speed = (26+14+15+17)/2 km/h
=36 km/h
=36x1000/(60x60) m/s
=10 m/s 
(a) For the average speed, divide the total distance travelled, in ANY direction, by 2 hours. That would result in:
(26 + 14 + 15 + 17)/2 = 36 km/h = 36000 m/3600 s = 10.0 m/s
(b) For the average velocity, divide the distance between the starting and finishing points (a vector sum) by 2 hours.
The vector velocity from start to finish is
(26i + 14j 10.607i + 10.607j 17i)/2
= (1.607i 24.607j)/2
(i represents the east direction and j represents the north direction)
= 0.803 i 12.304 j
The magnitude of the average velocity (average speed) is 12.33 km/h, and the direction is slightly north of west. 
Simple question
:D
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