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

Mar 20, 2018

The average speed is $= 6.36 m {s}^{-} 1$ and the average velocity is $= 1.27 m {s}^{-} 1$

#### Explanation:

The total distance travelled is

$d = \left(7 \cdot 4\right) + \left(6 \cdot 7\right) = 28 + 42 = 70 m$

The total time is

$t = 4 + 7 = 11 s$

The average speed is

$\overline{v} = \frac{d}{t} = \frac{70}{11} = 6.36 m {s}^{-} 1$

The apparent distance travelled is

${d}_{1} = 42 - 28 = 14 m$

The average velocity is

${v}_{1} = {d}_{1} / t = \frac{14}{11} = 1.27 m {s}^{-} 1$

Mar 20, 2018

Interpreting the question to be asking for 'displacement'

#### Explanation:

Let the Northern vector be positive.
So the Southern vector is negative.

Let the resultant be $R$

Thus we have:

The displacement (resultant)$R = \left(7 \frac{m}{\cancel{s}} \times 4 \cancel{s}\right) - \left(6 \frac{m}{\cancel{s}} \times 7 \cancel{s}\right)$

$R = - 14 m$ which is south as it is negative

So a better representation is:

The total time for this displacement to be occurring is $\left(4 + 7\right)$ seconds.

Note that in how I set this up a negative value means south from the starting point.

So the average $\underline{\text{rate of displacement}}$ is $- \frac{14}{11}$ Exact value

In decimal form this is $- 1.27 \overline{27}$ where the bar means that the 27 repeats for ever,
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Average speed:

$\left[\left(7 \frac{m}{\cancel{s}} \times 4 \cancel{s}\right) + \left(6 \frac{m}{\cancel{s}} \times 7 \cancel{s}\right)\right] \times \frac{1}{4 s + 7 s} = \frac{70}{11} \frac{m}{s}$

Average speed $6.45 \overline{45}$ m/s

Mar 20, 2018

Average speed $\text{= 6.36 m/s}$

Average velocity $\text{= 1.27 m/s along South}$

#### 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 >> = (("7 m/s × 4 s") + ("6 m/s × 7 s"))/("4 s + 7 s") = color(blue)"6.36 m/s"

• Let unit vector along North be $\hat{j}$. So, unit vector along South will be $- \hat{j}$

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

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

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

$\left\langle\vec{V}\right\rangle = \frac{28 \hat{j} - 42 \hat{j}}{11} \setminus \text{m/s}$

$\left\langle\vec{V}\right\rangle = \frac{- 14 \hat{j}}{11} \setminus \text{m/s" = -1.27hatj\ "m/s" = color(blue)"1.27 m/s along South}$