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?

3 Answers
Mar 20, 2018

Answer:

The average speed is #=6.36ms^-1# and the average velocity is #=1.27ms^-1#

Explanation:

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#

Mar 20, 2018

Answer:

Interpreting the question to be asking for 'displacement'

Explanation:

Tony B

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:
Tony B

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

Mar 20, 2018

Answer:

Average speed #"= 6.36 m/s"#

Average velocity #"= 1.27 m/s along South"#

Explanation:

#"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"#