Answer :
To solve the problem, we need to determine the distance and displacement for each object and check if the distance is three times the displacement for any of them.
1. Object W:
- Motion: 3 units left, then 3 units right.
- Distance calculation: [tex]\(3 + 3 = 6\)[/tex] units.
- Displacement calculation: [tex]\(-3 + 3 = 0\)[/tex] units.
- Does [tex]\(6 = 3 \times 0\)[/tex]? No.
2. Object X:
- Motion: 6 units right, then 18 units right.
- Distance calculation: [tex]\(6 + 18 = 24\)[/tex] units.
- Displacement calculation: [tex]\(6 + 18 = 24\)[/tex] units.
- Does [tex]\(24 = 3 \times 24\)[/tex]? No.
3. Object Y:
- Motion: 8 units left, then 24 units right.
- Distance calculation: [tex]\(8 + 24 = 32\)[/tex] units.
- Displacement calculation: [tex]\(-8 + 24 = 16\)[/tex] units.
- Does [tex]\(32 = 3 \times 16\)[/tex]? No.
4. Object Z:
- Motion: 16 units right, then 8 units left.
- Distance calculation: [tex]\(16 + 8 = 24\)[/tex] units.
- Displacement calculation: [tex]\(16 - 8 = 8\)[/tex] units.
- Does [tex]\(24 = 3 \times 8\)[/tex]? Yes.
Therefore, the object that has a distance three times as great as its displacement is [tex]\( \boxed{Z} \)[/tex].
1. Object W:
- Motion: 3 units left, then 3 units right.
- Distance calculation: [tex]\(3 + 3 = 6\)[/tex] units.
- Displacement calculation: [tex]\(-3 + 3 = 0\)[/tex] units.
- Does [tex]\(6 = 3 \times 0\)[/tex]? No.
2. Object X:
- Motion: 6 units right, then 18 units right.
- Distance calculation: [tex]\(6 + 18 = 24\)[/tex] units.
- Displacement calculation: [tex]\(6 + 18 = 24\)[/tex] units.
- Does [tex]\(24 = 3 \times 24\)[/tex]? No.
3. Object Y:
- Motion: 8 units left, then 24 units right.
- Distance calculation: [tex]\(8 + 24 = 32\)[/tex] units.
- Displacement calculation: [tex]\(-8 + 24 = 16\)[/tex] units.
- Does [tex]\(32 = 3 \times 16\)[/tex]? No.
4. Object Z:
- Motion: 16 units right, then 8 units left.
- Distance calculation: [tex]\(16 + 8 = 24\)[/tex] units.
- Displacement calculation: [tex]\(16 - 8 = 8\)[/tex] units.
- Does [tex]\(24 = 3 \times 8\)[/tex]? Yes.
Therefore, the object that has a distance three times as great as its displacement is [tex]\( \boxed{Z} \)[/tex].