Question

one of these representations is not like the others. tap on the one that doesnt belong.

arrows
represent
acceleration
vectors

the objects
vertical velocity
changes by the
same amount
every second.

Explanation:

Step1: Analyze the first table (displacement)

In projectile - motion (assuming this is a projectile - like situation), the horizontal displacement \(d_x = v_{0x}t\) (constant - velocity motion in the \(x\) - direction if there is no air resistance) and the vertical displacement \(d_y=v_{0y}t-\frac{1}{2}gt^{2}\). For \(d_x\), when \(t = 1s\), \(d_x=8m\), \(v_{0x}=8m/s\); when \(t = 2s\), \(d_x = 16m\) (since \(d_x=v_{0x}t\) with \(v_{0x}=8m/s\)). For \(d_y\), using \(d_y=v_{0y}t-\frac{1}{2}gt^{2}\), if \(v_{0y} = 0\), \(d_y=-\frac{1}{2}(9.8)t^{2}\). When \(t = 1s\), \(d_y=- 4.9m\); when \(t = 2s\), \(d_y=-19.6m\) (\(d_y=-\frac{1}{2}(9.8)(2)^{2}=-19.6m\))

Step2: Analyze the second table (velocity)

In projectile - motion, the horizontal velocity \(v_x\) is constant (\(v_x = v_{0x}\) if there is no air resistance) and the vertical velocity \(v_y=v_{0y}-gt\). If \(v_{0x}=8m/s\), \(v_x\) is constant. But if \(v_{0y}=-9.8m/s\), then at \(t = 1s\), \(v_y=-9.8 - 9.8=-19.6m/s\); at \(t = 2s\), \(v_y=-9.8-9.8\times2=-29.4m/s\). The given velocity table has \(v_y=-9.8m/s\) for all \(t\), which is incorrect.

Step3: Analyze the graph

The graph with acceleration vectors (assuming vertical acceleration due to gravity \(a=-g=-9.8m/s^{2}\) and constant horizontal velocity) is a valid representation of projectile - motion.

Step4: Analyze the text

The text “The object's vertical velocity changes by the same amount every second” is correct. Since \(a_y=\frac{\Delta v_y}{\Delta t}=-g\), \(\Delta v_y=-g\Delta t\), so the change in vertical velocity per second is \(\Delta v_y=-9.8m/s\) per second.

Answer:

The table with \(v_x\) and \(v_y\) values (the bottom - right table) doesn't belong.