To determine which object's distance is three times as great as its displacement, we first need to calculate the distance and displacement for each object, and then compare them.
Object W:
- Motion: 3 units left, then 3 units right
- Distance: The total path travelled by the object.
- Distance = 3 (left) + 3 (right) = 6 units.
- Displacement: The net change in position from the starting point.
- Displacement = 3 (right) - 3 (left) = 0 units.
- Net displacement is simply the final position minus the initial position, which is 0.
Object X:
- Motion: 6 units right, then 18 units right
- Distance:
- Distance = 6 (right) + 18 (right) = 24 units.
- Displacement:
- Displacement = 6 (right) + 18 (right) = 24 units.
- Final displacement is 24 units towards the right.
Object Y:
- Motion: 8 units left, then 24 units right
- Distance:
- Distance = 8 (left) + 24 (right) = 32 units.
- Displacement:
- Displacement = 24 (right) - 8 (left) = 16 units.
- Net displacement is 16 units towards the right.
Object Z:
- Motion: 16 units right, then 8 units left
- Distance:
- Distance = 16 (right) + 8 (left) = 24 units.
- Displacement:
- Displacement = 16 (right) - 8 (left) = 8 units.
- Net displacement is 8 units towards the right.
Now, we need to determine which of these objects has a distance that is three times its displacement.
- For W: Distance = 6 units, Displacement = 0 units (undefined ratio since displacement is zero).
- For X: Distance = 24 units, Displacement = 24 units.
- For Y: Distance = 32 units, Displacement = 16 units, which gives us a ratio of [tex]\(\frac{32}{16} = 2\)[/tex] (not equal to 3).
- For Z: Distance = 24 units, Displacement = 8 units, which gives us a ratio of [tex]\(\frac{24}{8} = 3\)[/tex].
Thus, the object that has a distance three times as great as its displacement is Object Z.
