Jose Buys A New Moped. He Travels 3km South And Then 4km East. How Far Does He Need To Go To Get Back To Where He Started?

Jose buys a new moped. He travels 3km south and then 4km east. How far does he need to go to get back to where he started?

 

To get back to its original position, it can either be in an opposite direction of the total distance travelled  (much longer) or opposite direction of the displacement (shorter than the total distance travelled).

For the total distance travelled:

3 km + 4 km = 7 km

Thus, Jose needs to go back 7 km in an opposite direction (4 km West and 3 km North).

For the displacement:

Let d represent as displacement.

Let R represent as resultant.

Let Φ represents as the angle.

Given:

d1 = 3 km, 270°

d2 = 4 km, 0°

Formula:

For horizontal x-axis:

(n)(cosΦ)

For vertical y-axis:

(n)(sinΦ)

For the resultant (or the magnitude displacement):

R = √((x)^2 + (y)^2)

For the angle (althought unnecessary):

Φ = arctan (y/x)

Equation:

d1x = (3 km)(cos270) = 0 km

d1y = (3 km)(sin270) = -3 km

d2x = (4 km)(cos0) = 4 km

d2y = (4 km)(sin0) = 0 km

Summation of dx = (d1 + d2)x = 4 km

Summation of dy = (d1 + d2)y = -3 km

R = 5 km

Φ = 143.1301024°

Thus, Jose needs to go back 5 km at an angle of approximately 143.13° (or 143°).

Therefore, the magnitude of the displacement (5 km) is more convenient than using the distance travelled (7 km).

(NOTE: The illustrated angle of the vector diagram is inaccurate; but, its illustration is precise.)