Question

20 POINTS!!!

Answer 1

a) Average velocity for her run north: 2.68 m/s north

b) The average velocity for the whole trip is zero

c) The average speed for the whole trip is 1.95 m/s

Explanation:

a)

In this part of the problem we only want to consider the part when Micha is running north.

The average velocity is given by:

$v=\frac{d}{t}$

where

d is the displacement (a vector connecting the initial position to the final position of motion)

t is the time

For the part running north, we have

$d=2 mil \cdot 1609 = 3218 m$ north is the displacement

$t = 20 min \cdot 60 = 1200 s$ is the time interval

Substituting,

$v=\frac{3218}{1200}=2.68 m/s$ north (we have to specify also the direction, since velocity is a vector)

b)

As we said, average velocity is given by

$v=\frac{d}{t}$

where

d is the displacement

t is the time

If we consider the whole trip, however, the displacement is zero:

d = 0

Because Micha returns home, so the initial position corresponds to the final position of motion. Therefore, the average velocity is also zero:

v = 0

c)

The average speed is given by

$s=\frac{d}{t}$

where

d is the  distance covered (the total length of the path, regardless of the direction)

t is the time interval

Since MIcha ran 2 miles north and then 2 miles back, the total distance is

$d=2 mi + 2mi = 4 mi \cdot 1609 = 6436 m$

And the time taken is

$t=20 min + 15 min + 20 min = 55 min \cdot 60 = 3300 s$

So, the average speed is

$s=\frac{6436}{3300}=1.95 m/s$

And since speed is a scalar, there is no need to specify a direction.

Learn more about speed and velocity:

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