That's a peculiar question. I'd call that the 'effective average speed',
or the 'apparent average speed'.
displacement+time divided by distance
(Angular displacement divided by 360 degrees or pi radians)/(Time measured in minutes)
The magnitude of displacement is equal to distance traveled when motion is in a straight line.
Displacement over time is velocity - distance over time is speed.The reason is that:Displacement is a vector quantity (has a direction as well as a magnitude), whereas speed is a scalar quantity (has no direction, only magnitude).
Velocity is displacement divided by time. Displacement is different from distance traveled, as displacement states how far you traveled in RELATION to a starting point. The formula for Velocity is ---- v = x / t v = Velocity x = Displacement t = Time velocity is a vector quantity so the direction should also be specified unless it is implicit in the problem. ----
Displacement divided by time will give you the motion of an object that has no unbalanced force acting on it
It's as simple as total displacement divided by total time. Be careful though. If this is a velocity problem, displacement does not always equal distance.
Ifk
Average Velocity
no
Velocity is displacement/time.
a=dv/dt average velocity = displacement divided by time take. so average velocity = displacement/time taken.
displacement+time divided by distance
Every time the unicycle returns to its starting point, the average velocity equals zero. C. The total displacement divided by the time.
The Average Velocity on a position time graph or a velocity time graph.
The total displacement divided by the time. The slope of the displacement vs. time graph.
The slope of the ant's displacement vs. time graph The total displacement divided by the time.