# An object travels North at  8 m/s for 3 s and then travels South at  7 m/s for  8 s. What are the object's average speed and velocity?

Apr 26, 2018

Average Speed

$\overline{v} \approx 7.27 \textcolor{w h i t e}{l} {\text{m"*"s}}^{- 1}$

Average Velocity

$\overline{\textsf{v}} \approx 5.54 \textcolor{w h i t e}{l} {\text{m"*"s}}^{- 1}$

#### Explanation:

$\text{Speed}$ equals distance over time whereas
$\text{Velocity}$ equals displacement over time.

Total distance travelled- which is independent of the direction of motion- in $3 + 8 = 11 \textcolor{w h i t e}{l} \text{seconds}$

$\Delta s = {s}_{1} + {s}_{2} = {v}_{1} \cdot {t}_{1} + {v}_{2} \cdot {t}_{2} = 8 \cdot 3 + 7 \cdot 8 = 80 \textcolor{w h i t e}{l} \text{m}$

Average speed

bar(v)=(Delta s)/(Delta t)=(80color(white)(l)"m")/(11color(white)(l)"s")~~7.27color(white)(l)"m"*"s"^(-1)

The two components of the final displacement, ${\textsf{x}}_{1}$ and ${\textsf{x}}_{2}$, are normal (a.k.a., perpendicular ) to each other.

Hence, directly apply the Pythagorean theorem to find the
displacement from the initial position after $11 \textcolor{w h i t e}{l} \text{seconds}$

$\Delta \textsf{x} = \sqrt{{\textsf{x}}_{1}^{2} + {\textsf{x}}_{2}^{2}} = \sqrt{{\left({\textsf{v}}_{1} \cdot {t}_{1}\right)}^{2} + {\left({\textsf{v}}_{2} \cdot {t}_{2}\right)}^{2}}$

$= \sqrt{{\left(8 \cdot 3\right)}^{2} + {\left(7 \cdot 8\right)}^{2}} = 8 \sqrt{58} \textcolor{w h i t e}{l} \text{m}$

Average velocity

bar(sf(v))=(Delta sf(x))/(Delta t)=(8sqrt(58)color(white)(l)"m")/(11color(white)(l)"s")~~5.54color(white)(l)"m"*"s"^(-1)