Question

A protester carries his sign of protest, starting from the origin of an xyz coordinate system, with the xy plane horizontal. He moves 40 m in the negative direction of the x axis, then 20 m along a perpendicular path to his left, and then 25 m up a water tower.
(a) In unit-vector notation, what is the displacement of the sign from start to end?
(b) The sign then falls to the foot of the tower. What is the magnitude of the displacement of the sign from start to this new end?

Answer 1

Answer:

Part A:

$Displacement\ vector= -40\hat i+20\hat j+25\hat k$

Part B:

Magnitude of displacement=44.721359 m

Explanation:

Given Data:

40 m in negative direction of the x axis

20 m perpendicular path to his left (considering +ve y direction)

25 m up a tower (Considering +ve z direction)

Required:

• In unit-vector notation, what is the displacement of the sign from start to end.
• The magnitude of the displacement of the sign from start to this new end=?

Solution:

Part A:

$Displacement= (0-40)\hat i+(0+20)\hat j+(0+25)\hat k\\Displacement= -40\hat i+20\hat j+25\hat k$

where:

i,j,k are unit vectors

Part B:

Sign falls to foot of tower so z=0

$Displacement= -40\hat i+20\hat j+0\hat k$

Magnitude of displacement:

$Magnitude\ of\ displacement=\sqrt{(-40)^2+(20)^2+0^2} \\Magnitude\ of\ displacement=44.721359\ m$