An ant on a picnic table travels 30 cm eastward, then 15 cm northward, then 20 cm westward, and finally 15 cm southward. Show that the total distance traveled is 65 cm, but the displacement is 10 cm. Complete your work on a separate sheet, take a photo of your work, and upload/paste it into the text box.



Answer :

Solution:

Let's break down the problem into two parts: the total distance traveled and the displacement.

### Total Distance Traveled

The ant travels in four different segments:
1. Eastward: 30 cm
2. Northward: 15 cm
3. Westward: 20 cm
4. Southward: 15 cm

To find the total distance traveled, we simply add up all the distances of these segments:

[tex]\[ \text{Total Distance} = 30 \text{ cm} + 15 \text{ cm} + 20 \text{ cm} + 15 \text{ cm} \][/tex]

[tex]\[ \text{Total Distance} = 80 \text{ cm} \][/tex]

So, the total distance traveled by the ant is 80 cm.

### Displacement

Displacement is the straight-line distance from the starting point to the ending point. To find this, we need to consider the net movement in the east-west direction and the net movement in the north-south direction.

1. Net East-West Movement:
- The ant moves 30 cm eastward and then 20 cm westward.
- Net East-West Movement = 30 cm (east) - 20 cm (west) = 10 cm eastward.

2. Net North-South Movement:
- The ant moves 15 cm northward and then 15 cm southward.
- Net North-South Movement = 15 cm (north) - 15 cm (south) = 0 cm.

Now, we find the displacement using the Pythagorean theorem. Since the net north-south movement is zero, the displacement is effectively just the net east-west movement.

[tex]\[ \text{Displacement} = \sqrt{(10 \text{ cm})^2 + (0 \text{ cm})^2} \][/tex]

[tex]\[ \text{Displacement} = \sqrt{100} \][/tex]

[tex]\[ \text{Displacement} = 10 \text{ cm} \][/tex]

So, the displacement of the ant is 10 cm.

In conclusion:
- The total distance traveled by the ant is 80 cm.
- The displacement of the ant is 10 cm.

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