Question

A runner in a relay race runs 20 m north, turns around and runs south for 30 m, then turns north again and runs 40 m. The entire run took 30 seconds. What was the average speed of the runner? What was the average velocity of the runner?

Answer 1

Answer:

• The average speed its 1 m/s
• The average velocity its 1 m/s to the north.

Explanation:

So, lets say the runner stars from the position $x_0$. Lets make this point the origin of a coordinate system in which the vector i points north.

$x_0 = (0,0)$

Now, in the first sections of the race, he runs 20 meters north, so, he finds himself at:

$x_1 = x_0 + 20 m * i = (0,0) \ + (20 \ m,0)$.

$x_1 = (20 \ m,0)$.

The, he runs 30 meters south

$x_2 = x_1 - 30 \ m * i = (20 \ m,0)-(30 \ m,0)$

$x_2 = (-10 \ m,0)$

Finally, he runs 40 meter north

$x_3 = x_2 + 40 \ m * i = (-10 \ m,0)+(40 \ m,0)$

$x_3 = (30 \ m,0)$.

This is our displacement vector. Now, the average speed will be:

$\frac{distance}{time}$.

The distance its the length of the displacement vector,

$d=\sqrt{x^2+y^2}$

$d=\sqrt{(30 \ m)^2+0^2}$

$d=30 \ m$

So, the average speed its:

$\frac{30 \ m }{30 \ s} = 1\frac{m}{s}$.

The average velocity, instead, its:

$\vec{v} = \frac{displacement}{time}$

$\vec{v} = \frac{(30 \ m ,\ 0)}{30 \ s}$

$\vec{v} = (1 \ \frac{m}{s} ,\ 0)$

This is, 1 m/s north.