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?

The total distance travelled is

`d=(7*4)+(6*7)=28+42=70m`

The total time is

`t=4+7=11s`

The average speed is

`barv=d/t=70/11=6.36ms^-1`

The apparent distance travelled is

`d_1=42-28=14m`

The average velocity is

`v_1=d_1/t=14/11=1.27ms^-1`

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

Let the resultant be `R`

Thus we have:

The displacement (resultant)`R =( 7 m/cancel(s)xx4 cancel(s))-(6 m/cancel(s) xx7cancel(s))`

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

So a better representation is:

The total time for this displacement to be occurring is `(4+7)` seconds.

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

So the average `ul("rate of displacement")` is `-14/11` Exact value

In decimal form this is `-1.27bar27` where the bar means that the 27 repeats for ever,
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Average speed:

`[( 7 m/cancel(s)xx4 cancel(s))+(6 m/cancel(s) xx7cancel(s))]xx1/(4s+7s) = 70/11 m/s`

Average speed `6.45bar(45)` m/s

`"Average speed" = "Total distance"/"Total time"`

`<< V >> = (v_1t_1 + v_2t_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 `hatj`. So, unit vector along South will be `-hatj`

`"Average velocity" = "Total displacement"/"Total time"`

`<< vecV >> = (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")`

`<< vecV >> = (28hatj - 42hatj)/11\ "m/s"`

`<< vecV >> = (-14 hatj)/11\ "m/s" = -1.27hatj\ "m/s" = color(blue)"1.27 m/s along South"`