Question

A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastward. Carrying a bad penny, he walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high

(a) In unit-vector notation, what is the displacement of the penny from start to its landing point?
(b) When the man returns to the origin, what is the magnitude of his displacement for the return trip?

Answer 1

a) Displacement of penny = 1300 i + 2400 j - 640 k

b) Magnitude of his displacement = 2729.47 m

Explanation:

a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.

1300 m east = 1300 i

2400 m north = 2400 j

Drops the penny from a cliff 640 m high = -640 k

Displacement of penny = 1300 i + 2400 j - 640 k

b) Displacement of man for return trip = -1300 i - 2400 j

    $\texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m$

    Magnitude of his displacement = 2729.47 m

Answer 2

Answer:

Explanation:

d1 = 1300 m east

d2 = 2400 m north

d3 = 640 m downward

(a)

The displacement of penny is given by

$\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}$

$\overrightarrow{d}=1300\widehat{i}+2400 \widehat{j}-640\widehat{k}$

(b) For the return journey of man, the displacement is given by

$\overrightarrow{d}=-\overrightarrow{d_{1}}-\overrightarrow{d_{2}}$

$\overrightarrow{d}=-1300\widehat{i}-2400 \widehat{j}$

The magnitude of the displacement is given by

$d=\sqrt{1300^{2}+2400^{2}}=2729.47 m$