# An object travels North at  8 m/s for 6 s and then travels South at  7 m/s for  4 s. What are the object's average speed and velocity?

Mar 12, 2018

Average speed $\text{= 7.6 m/s}$

Average velocity $\text{= 2 m/s along North}$

#### Explanation:

$\text{Average speed" = "Total distance"/"Total time}$

$\left\langleV\right\rangle = \frac{{v}_{1} {t}_{1} + {v}_{2} {t}_{2}}{{t}_{1} + {t}_{2}}$

<< V >> = (("8 m/s × 6 s") + ("7 m/s × 4 s"))/("6 s + 4 s") = color(blue)"7.6 m/s"

$\text{Average velocity" = "Total displacement"/"Total time}$

$\left\langle\vec{V}\right\rangle = \frac{\vec{{v}_{1}} {t}_{1} + \vec{{v}_{2}} {t}_{2}}{{t}_{1} + {t}_{2}}$

<< vecV >> = (("8 m/s" (hatj) × "6 s") + ("7 m/s" (-hatj) × "4 s"))/("6 s + 4 s")

$\left\langle\vec{V}\right\rangle = \frac{48 \hat{j} - 28 \hat{j}}{10} \text{m/s}$

$\left\langle\vec{V}\right\rangle = \frac{20 \hat{j}}{10} \text{m/s" = 2hatj "m/s" = color(blue)"2 m/s along North}$

Mar 12, 2018

$\text{Average speed} = 7.6 \frac{m}{s}$ and $\text{Average velocity} = 2.0 \frac{m}{s}$

#### Explanation:

The distance North, ${s}_{n}$, it traveled during that first part was

${s}_{n} = {v}_{n} \cdot {t}_{n} = 8 \frac{m}{s} \cdot 6 s = 48 m$

The distance Sorth, ${s}_{s}$, it traveled during that first part was

${s}_{s} = {v}_{s} \cdot {t}_{s} = 7 \frac{m}{s} \cdot 4 s = 28 m$

Average speed

$\text{Average speed " = "total distance"/"total time" " }$ So, plugging in our data,

$\text{Average speed } = \frac{48 m + 28 m}{6 s + 4 s} = \frac{76 m}{10 s}$
$\text{Average speed} = 7.6 \frac{m}{s}$

Average velocity

$\text{Average velocity " = "total displacement"/"total time}$
To determine displacement, we need a rule for what the positive direction is. I declare that North is the positive direction. So, plugging in our data,

$\text{Average velocity} = \frac{48 m - 28 m}{6 s + 4 s} = \frac{20 m}{10 s}$
$\text{Average velocity} = 2.0 \frac{m}{s}$

I hope this helps,
Steve