Question

An adventurous dog strays from home, runs three blocks east, two blocks north, one block east, one block north, and two blocks west. Assuming that each block is about 100 m, how far from home and in what direction is the dog? Use a graphical method.

a) 600 m east; 200 m north
b) 200 m west; 300 m south
c) 200 m east; 200 m north
d) 200 m east; 300 m north

Answer 1

b

Step-by-step explanation:

Answer 2

The dog's net displacement is 2 blocks east and 3 blocks north from home.

Step-by-step explanation:

To find the dog's net displacement, we can represent each block as a vector and add them together. Let's assign the +x-axis as east and the +y-axis as north.

The first block is 3 blocks east, which can be represented as a vector of magnitude 3 and in the +x direction.

The second block is 2 blocks north, which can be represented as a vector of magnitude 2 and in the +y direction.

The third block is 1 block east, which can be represented as a vector of magnitude 1 and in the +x direction.

The fourth block is 1 block north, which can be represented as a vector of magnitude 1 and in the +y direction.

The fifth block is 2 blocks west, which can be represented as a vector of magnitude 2 and in the -x direction.

To find the net displacement, we add all these vectors together. The x-components have a net sum of 3 + 1 - 2 = 2, and the y-components have a net sum of 2 + 1 = 3.

Therefore, the dog is 2 blocks east and 3 blocks north from home.