Asked by Anonymous
The ants of the North Creek Ant Colony traverse a straight trail between their colony and a nearby picnic table. One ant, returning after quenching its thirst with a portion of a tasty beverage dripping from the table, is first observed along the trail 3.1 m from the colony. Then, after traveling a total of 2.6 m while staggering back and forth along the trail, the ant falls unconscious 1.3 m from the colony exactly 43 s later.
Defining the entrance of the colony as the origin and the positive direction toward the picnic table, determine the ant's average velocity ð£avg during this time interval.
Defining the entrance of the colony as the origin and the positive direction toward the picnic table, determine the ant's average velocity ð£avg during this time interval.
Answers
Answered by
Anonymous
Well the average velocity is the displacement divided by the time
unlike the
average speed which is the distance divided by the time
displacement includes direction and the direction is negative, toward home from the table
so the displacement was - (3.1 - 1.3) = - 1.8
so velocity is -1.8 / 43 m/s = - 0.0232 m/s or -2.32 cm/s (almost -an inch a second)
unlike the
average speed which is the distance divided by the time
displacement includes direction and the direction is negative, toward home from the table
so the displacement was - (3.1 - 1.3) = - 1.8
so velocity is -1.8 / 43 m/s = - 0.0232 m/s or -2.32 cm/s (almost -an inch a second)
Answered by
oobleck
avg velocity = displacement/time = (1.3-3.1)/43 = -0.04186 m/s
the 2.6m traveled only affects the speed, not the velocity.
the 2.6m traveled only affects the speed, not the velocity.
Answered by
Anonymous
whoops, divided wrong, - 0.041860 m/s
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