Question

one of these representations is not like the others. tap on the one that doesnt belong. \the acceleration is a constant value and always directed downwards.\

Explanation:

Step1: Analyze the physical situation

This is a projectile - motion - related problem. In projectile motion, the horizontal velocity \(v_x\) is constant (assuming no air resistance), and the vertical acceleration \(a_y=-g = - 9.8\ m/s^{2}\) (constant). The vertical velocity \(v_y=v_{y0}+a_y t\), which is a linear function of time \(t\).

Step2: Analyze the table

In the table, the value of \(v_y=-9.8\ m/s\) for all \(t\) values. But according to the kinematic equation \(v_y = v_{y0}+a_y t\) (\(a_y=-9.8\ m/s^{2}\)), \(v_y\) should change with time. For example, if \(v_{y0}=-9.8\ m/s\) and \(a_y=-9.8\ m/s^{2}\), then at \(t = 1s\), \(v_y=v_{y0}+a_y t=-9.8+( - 9.8)\times1=-19.6\ m/s\).

Step3: Analyze the graphs

• The velocity - vector graph: In projectile motion, the horizontal component of velocity is constant (horizontal arrows have the same length), and the vertical component of velocity changes (vertical arrows change in length) due to the constant vertical acceleration \(a_y=-g\).
• The vertical - velocity - time graph: The slope of the \(v_y - t\) graph is \(a_y=-9.8\ m/s^{2}\), which is consistent with the kinematic equation \(v_y=v_{y0}+a_y t\).
• The horizontal - velocity - time graph: Since there is no acceleration in the horizontal direction (\(a_x = 0\)), \(v_x\) is constant, which is also consistent with the kinematic equation \(v_x=v_{x0}+a_x t=v_{x0}\) (constant).

Answer:

The table.