Question

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

1) The average velocity while running north is 6 mph

2) The average velocity for the whole trip is zero

3) The average speed for the whole trip is 4.4 mph

Explanation:

1)

The average velocity is defined as:

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

where

d is the displacement

t is the time interval

Here we want to analyze the motion of Micha while she runs north only. The displacement is the distance in a straight line between the initial position and final position, so

$d=2 miles$

while the time interval is

$t=20 min = \frac{1}{3}h$

So, the average velocity is

$v=\frac{2}{1/3}=6 mph$

and the direction is the same as the displacement (north).

2)

As we said, the displacement is the distance in a straight line between the initial position and final position of motion.

If we consider the whole trip, the final position of the Micha is equal to her initial position, since she walked back to home: therefore,

displacement = d = 0

And so, the average velocity is also zero:

$v=0$

3)

The average speed is defined as

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

where

d is the distance covered

t is the time interval considered

The distance covered is just the length of the total path covered regardless of the direction, so for the entire jog we have:

d = 2 mi + 2 mi = 4 miles

(since she runs 2 miles north and then another 2 miles back)

The total time is

$t=20 min + 15 min + 20 min = 55 min = \frac{55}{60}h = \frac{11}{12}h$

So, the average speed is

$v=\frac{4}{11/12}=\frac{48}{11} mph = 4.4 mph$

Learn more about speed and velocity:

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