To find which object has a distance that is three times as great as its displacement, we'll calculate the distance and displacement for each object.
### Object W
- Motion: 3 units left, then 3 units right
- Distance traveled:
[tex]\[
\text{Distance} = |3| + |3| = 3 + 3 = 6 \, \text{units}
\][/tex]
- Displacement:
[tex]\[
\text{Displacement} = -3 + 3 = 0 \, \text{units}
\][/tex]
Since the displacement is 0, it's impossible for any ratio involving displacement to be valid here.
### Object X
- Motion: 6 units right, then 18 units right
- Distance traveled:
[tex]\[
\text{Distance} = |6| + |18| = 6 + 18 = 24 \, \text{units}
\][/tex]
- Displacement:
[tex]\[
\text{Displacement} = 6 + 18 = 24 \, \text{units}
\][/tex]
[tex]\[
\text{Ratio} = \frac{\text{Distance}}{\text{Displacement}} = \frac{24}{24} = 1
\][/tex]
This does not meet the condition of being three times.
### Object Y
- Motion: 8 units left, then 24 units right
- Distance traveled:
[tex]\[
\text{Distance} = |8| + |24| = 8 + 24 = 32 \, \text{units}
\][/tex]
- Displacement:
[tex]\[
\text{Displacement} = -8 + 24 = 16 \, \text{units}
\][/tex]
[tex]\[
\text{Ratio} = \frac{\text{Distance}}{\text{Displacement}} = \frac{32}{16} = 2
\][/tex]
This does not meet the condition of being three times.
### Object Z
- Motion: 16 units right, then 8 units left
- Distance traveled:
[tex]\[
\text{Distance} = |16| + |8| = 16 + 8 = 24 \, \text{units}
\][/tex]
- Displacement:
[tex]\[
\text{Displacement} = 16 - 8 = 8 \, \text{units}
\][/tex]
[tex]\[
\text{Ratio} = \frac{\text{Distance}}{\text{Displacement}} = \frac{24}{8} = 3
\][/tex]
Since the ratio of distance to displacement for object Z is exactly 3, the answer is:
Z.
