Answer:
The answers to the question are
(a) The displacement vector is 146246.212 i + 146246.212 j
Magnitude of the displacement = 151052.536 m
Displacement direction is 37.46 ° North of East.
(b) The average velocity is
= 27.08 i + 23.343 j
The magnitude of the average velocity is 35.75 m/s
The direction of average velocity is 49.239 ° North of East.
Explanation:
We note the displacement and the direction for each of the stage as follows
Travel direction Travel velocity Duration Displacement
North 21.0 m/s 30.0 min 37800 m
East 30.0 m/s 40.0 min 72000 m
North east 35.0 m/s 50.0 min 105000 m
We note that the total distance moved north is given by
37800 + 105000×sin(45) = 37800 m + 74246.2 m = 112046.212 m
The total distance moved east is given by
72000 m + 74246.2 m = 146246.212 m
The displacement vector is given as
Δr = r₂ - r₁ = (x₂ - x₁)i + (y₂ - y₁)j + (z₂ - z₁)k
= 146246.212 i + 146246.212 j
Therefore the magnitude of the displacement is given by
[tex]\sqrt{x^2 +y^2} = \sqrt{112046.212^2+146246.212^2}[/tex] = 151052.536 m
The direction is given by tan⁻¹ [tex]\frac{112046.212 }{146246.212}[/tex] = 37.46 ° North of East
(b) The average velocity vector is given by knowing the displacement vector thus
Displacement vector =Δr = r₂ - r₁ = (x₂ - x₁)i + (y₂ - y₁)j + (z₂ - z₁)k
and [tex]v_{av} = \frac{r_2-r_1}{t_2-t_1}[/tex] that is vₐ = Δr/Δt
Therefore if we take our east to be the x axis, and the north to be our y axis, we have
Total displacement east is given above as 146246.212 m
Time taken to move east = 40+50 = 90 min = 5400 s
Average velocity in the x direction = 27.08 m/s
Total displacement North is given above as 112046.212 m
Time taken to move east = 30+50 = 80 min = 4800 s
Average velocity in the x direction = 23.343 m/s
The average velocity therefore is
v(average) = 27.08 i + 23.343 j
Magnitude of the average velocity = 35.75 m/s
The direction of average velocity is given by tan⁻¹[tex](\frac{27.08}{23.343})[/tex] = 49.239 ° North of East.
