A person walks 3km east in 9 minutes, then 4 km north in 28 minutes. There displacement is 5 km East 53 degrees North. What is the average velocity (in ms-1)? Thanks
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The average velocity is a vector quantity, which is the displacement over the time interval it takes for the displacement to happen.
The person walked 3km + 4km, and that's 7 km, but this is the distance, and distance is SPEED, not velocity. The displacement is a straight line from the initial position from the final position, which is 5km in this case. The time interval is 9 min + 28 min = 37 min though.
We want the velocity in meters per second, so let's begin by converting 5 km to meters and 37 min to seconds:
5 km * (1000 m/1 km) = 5000 m
37 min * (60 s/1 min) = 2220 s
Vavg. = Δx/Δt = 5000 m/2220 s = 2.2522522... m/s ≈ 2.26 m/s = 2.26 m s⁻¹
The person walked 3km + 4km, and that's 7 km, but this is the distance, and distance is SPEED, not velocity. The displacement is a straight line from the initial position from the final position, which is 5km in this case. The time interval is 9 min + 28 min = 37 min though.
We want the velocity in meters per second, so let's begin by converting 5 km to meters and 37 min to seconds:
5 km * (1000 m/1 km) = 5000 m
37 min * (60 s/1 min) = 2220 s
Vavg. = Δx/Δt = 5000 m/2220 s = 2.2522522... m/s ≈ 2.26 m/s = 2.26 m s⁻¹
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They are testing whether you understand the difference between average velocity (which is displacement / time) and average speed (total distance traveled / time).
5 km / 37 min = 5000 m / 2220 s = 2.25 m/s
5 km / 37 min = 5000 m / 2220 s = 2.25 m/s