1) Distance: 9 km, displacement: 6.7 km at [tex]63.4^{\circ}[/tex] north of east
2) Speed: 3 km/h, velocity: 2.2 km/h at [tex]63.4^{\circ}[/tex] north of east
Explanation:
1)
In this problem, the person travels:
5 km north
3 km east
1 km north
Therefore, the distance covered is simply the sum of these paths:
d = 5 + 3 +1 = 9 km
Instead, for the displacement we have to see what is the final position. The total displacement in the north direction is
[tex]d_y = 5 + 1 = 6 km[/tex]
While the displacement in the east direction is
[tex]d_x = 3 km[/tex]
Since the two are perpendicular in direction, we can find the length of the total displacement by using Pythagorean's theorem:
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{3^2+6^2}=6.7 km[/tex]
And the direction is
[tex]\theta=tan^{-1}(\frac{d_y}{d_x})=tan^{-1}(\frac{6}{3})=63.4^{\circ}[/tex] north of east.
2)
The average speed of a body is a scalar quantity given by
[tex]speed=\frac{distance}{time}[/tex]
In this problem,
distance = 9 km
time elapsed = 3 h
Therefore, the speed is
[tex]speed=\frac{9 km}{3 h}=3 km/h[/tex]
The average velocity of a body is a vector quantity given by
[tex]velocity = \frac{displacement}{time}[/tex]
In this problem,
displacement = 6.7 km
time elapsed = 3 h
Therefore, the velocity is
[tex]velocity = \frac{6.7}{3}=2.2 km/h[/tex]
And the direction is the same as the displacement ([tex]63.4^{\circ}[/tex] north of east)
Learn more about distance, displacement, speed and velocity:
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