An ant on a picnic table travels 40 cm eastward, then 25 cm northward, and finally 15 cm westward. What is the ant's directional displacement relative to its original position?
X = 40-15 = 25 cm.
Y = 25 cm.
tanA = Y/X = 25/25 = 1.00
A = 45o. = Direction.
Mag. = X / cos45 = 25/cos45 = 35.36 cm.
To determine the ant's directional displacement, we need to calculate the net displacement, which is the straight-line distance between the ant's starting point and ending point.
To begin, let's visualize the problem:
```
Starting point (0,0)
|
| (40 cm eastward)
|
-------Ant-------
|
| (25 cm northward)
|
|
|
| (15 cm westward)
|
```
First, the ant moves 40 cm eastward. This means the ant has traveled 40 cm in the positive x-direction and hasn't moved vertically (y-direction).
Next, the ant moves 25 cm northward. This implies an increase of 25 cm in the positive y-direction and no change in the x-direction.
Lastly, the ant moves 15 cm westward. This results in a decrease of 15 cm in the x-direction and no change in the y-direction.
Now, let's sum up the total displacement in both the x and y directions separately:
Net displacement in the x-direction = Total eastward displacement (40 cm) - Total westward displacement (15 cm) = 40 cm - 15 cm = 25 cm eastward
Net displacement in the y-direction = Total northward displacement (25 cm) = 25 cm
Thus, the ant's directional displacement, relative to its original position, is 25 cm eastward and 25 cm northward, or in vector notation:
Displacement vector = (25 cm, 25 cm).