Question

20 POINTS!!!
Micha ran 2 miles north in 20 minutes. She then turned back and ran 1 mile in 15 minutes. To cool down, she walked the rest of the way home. It took her 20 minutes.

What was Micha's average velocity for her run north?

What was Micha's average velocity for the whole trip?

What was her average speed for her entire jog?

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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