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Four different objects are placed on a number line at 0. The chart describes the motion of each object.

\begin{tabular}{|l|l|}
\hline
Object & Motion \\
\hline
W & 3 units left, then 3 units right \\
\hline
X & 6 units right, then 18 units right \\
\hline
Y & 8 units left, then 24 units right \\
\hline
Z & 16 units right, then 8 units left \\
\hline
\end{tabular}

Using the information in the chart, determine which object has a distance that is three times as great as its displacement.

A. W
B. X
C. Y
D. Z



Answer :

GinnyAnswer
To determine which of the objects has a distance that is three times as great as its displacement, we need to calculate both the distance traveled and the displacement for each object.

Object W:
1. Moves 3 units left and then 3 units right.
- Distance: [tex]\(3 + 3 = 6\)[/tex] units
- Displacement: [tex]\(-3 + 3 = 0\)[/tex] units

Object X:
1. Moves 6 units right and then 18 units right.
- Distance: [tex]\(6 + 18 = 24\)[/tex] units
- Displacement: [tex]\(6 + 18 = 24\)[/tex] units

Object Y:
1. Moves 8 units left and then 24 units right.
- Distance: [tex]\(8 + 24 = 32\)[/tex] units (distance always positive)
- Displacement: [tex]\(-8 + 24 = 16\)[/tex] units

Object Z:
1. Moves 16 units right and then 8 units left.
- Distance: [tex]\(16 + 8 = 24\)[/tex] units (distance always positive)
- Displacement: [tex]\(16 - 8 = 8\)[/tex] units

Now, we need to check which object has a distance that is three times its displacement:
- For Object W: Displacement is 0, so this cannot be checked because any number multiplied by 0 is still 0.
- For Object X: [tex]\(24 \neq 3 \times 24\)[/tex]
- For Object Y: [tex]\(32 \neq 3 \times 16\)[/tex]
- For Object Z: [tex]\(24 = 3 \times 8\)[/tex]

Therefore, Object [tex]\(Z\)[/tex] is the one whose distance is three times its displacement. The answer is [tex]\(Z\)[/tex].

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