Question 983054
It is not clearly specified,
but it seems like after going N for 6.0 minutes at 35 km/h,
the auto continues NE for 7.0 minutes at the same speed.
That is what I will assume.
The whole trip looks like this: {{{drawing(200,400,-0.5,3.5,-0.5,7.5,
circle(0,0,0.05),circle(2.887,3.72,0.05),
locate(0.1,0.2,start),locate(3,3.92,end),
triangle(0,3.5,2.887,3.5,2.887,6.387),
rectangle(2.887,3.5,2.737,3.65),
red(line(0,0,0,3.5)),red(arrow(0,0,0,3.5)),
green(line(0,3.5,2.887,6.387)),green(arrow(0,3.5,2.887,6.387)),
blue(line(2.887,6.387,2.887,3.72)),blue(arrow(2.887,6.387,2.887,3.72)),
locate(0.1,2,red(N)),locate(1.443,4.94,green(NE)),
locate(2.9,5.1,blue(S))
)}}}
The average velocity is a vector with magnitude and direction,
and can be found as the start-end displacement vector divided by the travel time.
The average speed is a scalar quantity (a number with units of km/h),
and can be found as the total distance covered along the trajectory divided by the travel time.
The distances traveled are as follows:
{{{red("N :")}}} {{{(35km/"60 minutes")*(6minutes)=3.5km}}}
{{{green("NE :")}}} {{{(35km/"60 minutes")*(7minutes)=49/12}}}{{{km=4.083km}}}(rounded)
{{{blue("S :")}}} {{{(20km/"60 minutes")*(8minutes)=8/3}}}{{{km=2.667km}}}(rounded).
Now the trips looks like this: {{{drawing(250,400,-0.8,4.2,-0.5,7.5,
circle(0,0,0.05),circle(2.887,3.72,0.05),
locate(0.1,0.2,start),locate(3,3.92,end),
triangle(0,3.5,2.887,3.5,2.887,6.387),
rectangle(2.887,3.5,2.737,3.65),
red(line(0,0,0,3.5)),red(arrow(0,0,0,3.5)),
green(line(0,3.5,2.887,6.387)),green(arrow(0,3.5,2.887,6.387)),
blue(line(2.887,6.387,2.887,3.72)),blue(arrow(2.887,6.387,2.887,3.72)),
locate(0.1,2,red(3.5km)),locate(1.443,4.94,green(4.083km)),
locate(2.95,5.1,blue(2.667km)),locate(1.34,3.5,x)
)}}}
The approximate total distance traveled is
{{{3.5km+4.083km+2.667km=10.250km}}} .
The total travel time is
{{{6minutes+7minutes+8minutes=21minutes}}} .
The average speed of an auto that travels a total of {{{10.250km}}} in {{{21minutes}}} is
{{{(10.250km/"21minutes")(60minutes/"1hour")=29.29}}}{{{"km / h"=29}}}{{{"km / h"}}} (rounded).
The eastward displacement is the horizontal leg of an isosceles right triangle with a hypotenuse of {{{49/12}}}{{{km=4.083km}}}(rounded).
It's measure in km is {{{x}}} and {{{2x^2=(49/12)^2=49^2/12^2}}} ,
so {{{x^2=49^2/(2*12^2)}}}--->{{{x=sqrt(2*49^2/(2^2*12^2))=sqrt(49^2/(2^2*12^2))*sqrt(2)=(49/(2*12))*sqrt(2)=49sqrt(2)/24=about2.8874}}} .
That is also the vertical leg of the same isosceles right triangle,
which is the northward displacement during the NE portion of the trip.
The total (approximate) northward displacement in km is
{{{3.5+2.887-2.667=3.72}}} .
Now the trips looks like this: {{{drawing(250,400,-1,4,-0.5,7.5,
circle(0,0,0.05),circle(2.887,3.72,0.05),
locate(0.1,0.2,start),locate(3,3.92,end),
triangle(0,3.5,2.887,3.5,2.887,6.387),
rectangle(2.887,3.5,2.737,3.65),
red(line(0,0,0,3.5)),green(line(0,3.5,2.887,6.387)),
blue(line(2.887,6.387,2.887,3.72)),
locate(0.1,2,red(3.5km)),locate(1.443,4.94,green(4.083km)),
locate(2.95,5.1,blue(2.667km)),locate(0.8,3.5,2.887km),
triangle(0,0,2.887,3.72,0,3.72),rectangle(0,3.72,0.15,3.57),
red(arc(0,0,2,2,-90,-52.2)),locate(0.2,0.95,red(theta)),
arrow(0,0,2.887,3.72),locate(1.45,1.86,displacement),
locate(-1,1.97,3.72km),arrow(-0.5,2,-0.5,3.72),arrow(-0.5,1.72,-0.5,0)
)}}}
The magnitude of the displacement (in km) is approximately
{{{sqrt(3.72^2+2.887^2)=4.709}}} .The 
The approximate) magnitude of the average velocity is
{{{(4.709km/"21minutes")(60minutes/"1hour")=13.45}}}{{{"km / h"=13.5}}}{{{"km / h"}}} (rounded),
and {{{red(theta)}}} gives us the direction.
{{{tan(red(theta))=2.887/3.72=0.7761}}}--->{{{red(theta)=37.8^o}}} ,
so the direction is {{{37.8^o}}} East of North.