8. A nervous cat, trying to cross a road, first moved 3.5 m east, then 4.2 m west, then 1.2 m
west, then 6.6 m east. What is the cat's
a) total displacement?
b) total distance?
) average velocity if the total time elapsed is 2 s.
d) average speed if the total time elapsed is 2 s.
a) The cat's total displacement is the vector sum of all its individual displacements. The positive direction is considered east, and the negative direction is west.
Total displacement = 3.5 m east - 4.2 m west - 1.2 m west + 6.6 m east
Total displacement = (3.5 m + 6.6 m) - (4.2 m + 1.2 m)
Total displacement = 10.1 m - 5.4 m
Total displacement = 4.7 m east
b) The total distance is the sum of all the magnitudes of the individual displacements.
Total distance = |3.5 m| + |4.2 m| + |1.2 m| + |6.6 m|
Total distance = 3.5 m + 4.2 m + 1.2 m + 6.6 m
Total distance = 15.5 m
c) The average velocity is defined as the total displacement divided by the total time elapsed.
Average velocity = Total displacement / Total time
Average velocity = 4.7 m east / 2 s
Average velocity = 2.35 m/s east
d) The average speed is defined as the total distance divided by the total time elapsed.
Average speed = Total distance / Total time
Average speed = 15.5 m / 2 s
Average speed = 7.75 m/s