A bug starts out at rest, 5 m to the right of the origin. It then starts moving on a trip. After 1 s, the bug is seen at 9 m to the right of the origin, travelling at 5 m/s to the right. After 5 s (from the start of the trip), the bug is seen at 2 m to the left of the origin, travelling at 7 m/s to the right. What is the average velocity of the bug for the entire trip?

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skyluke89 skyluke89 · Answer 1 · 2020-02-13T11:39:44+01:00

Average velocity: 0.4 m/s to the left

Explanation:

The average velocity of an object is equal to the ratio between the displacement of the object and the time taken:

[tex]v=\frac{d}{t}[/tex]

where

d is the displacement (the change in position of the object)

t is the time elapsed

For the bug in this problem, we have:

[tex]x_i = 0[/tex] (starting position)

Then at the end of the time interval (5 s), the final position of the bug is

[tex]x_f = -2 m[/tex] (to the left of the origin: we take left as negative direction)

So the displacement is

[tex]d=x_f - x_i = -2 -0 = -2 m[/tex] (left)

While the time interval of the entire trip is

[tex]t = 5 s[/tex]

Therefore, the average velocity is

[tex]v=\frac{-2}{5}=-0.4 m/s[/tex]

Which means 0.4 m/s to the left.

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