Answer:
scalar d = 54000 m
, v_average = 16,667 m / s
vector d = -4000 m
, moved 4000 to the west
v_average = 0
Explanation:
his is a uniform motion exercise, but we must be careful with quantities that are scalars and vector quantities
The distance traveled is a scalar
d = d₁ + d₂
d = 25000 + 29000
d = 54000 m
the speed is a scalar, in the exercise it is not specified if the speed of each trajectory or the average speed. Therefore we will calculate the two
v₁ = d₁ / t₁
v₁ = 25000/1500
v₁ = 16,667 m / s
v₂ = 29000/1740
v₂ = 16,667 m / s
Since the two speeds are equal, the average speed is
v = (v1 + v2) / 2
v_average = 16,667 m / s
now let's calculate the displacement that is a vector, so it has direction in addition to modules
suppose the eastward direction is positive and the bold are vectors
d = d₁ - d₂
d = 25000 - 29000
d = -4000 m
this means that it moved 4000 to the west
velocity is a vector, we assume positive eastward movement
v₁ = 16,667 m / s
v₂ = - 16,667 m / s
v_average = (v1 -v2) / 2
v_average = 0
