# A man walks 98m south. He stops and drops a letter into the mailbox, turns and walks 28m to the north. This trip took 4 minutes. What was the man's average speed for the walk?

Sep 12, 2015

17.5 meters per minute south.

#### Explanation:

Velocity is speed and direction. He would have traveled 70 meters south in four minutes. That's 17.5 meters per minute or 0.29 meters per second.

Sep 12, 2015

Average speed: $\text{31.5 m/min}$

#### Explanation:

This is a pretty straightforward displacement and velocity problem because all you have to do here is distinguish between average speed and average velocity.

Average speed will give you the ratio between the distance the man walked and the total time need, whereas average velocity will give you the ratio between the man's displacement and th etotal time of his trip.

To get the total distance the man walked during his trip, simply add the distance he walks south to the distance whe walked north

$d = {d}_{\text{south" + d_"north}}$

$d = \text{98 m" + "28 m" = "126 m}$

This means that his average speed was

$v = \frac{d}{t} = \text{126 m"/"4 min" = "31.5 m/min}$

You could convert this to meters per second if you wanted to

$31.5 \text{m"/color(red)(cancel(color(black)("min"))) * (1color(red)(cancel(color(black)("min"))))/"60 s" = "0.525 m/s}$

SIDE NOTE The man's average velocity will depend on his displacement, which is the distance between his starting point and finish point.

In this case, he walked 98 m south, then backtracked 28 m north, which means that he is now 70 m from where he first started.

Therefore, you have

$\overline{v} = \text{70 m"/"4 min" = "17.5 m/min}$