Question

one of these representations is not like the others. tap on the one that doesnt belong. arrows represent acceleration vectors arrows represent velocity vectors

Explanation:

Step1: Analyze the first graph (acceleration vectors)

In projectile motion, the acceleration is constant (\(a = g=- 9.8\ m/s^{2}\) downward). The first graph shows acceleration vectors (arrows) all in the same direction (downward), which is correct for projectile - like motion.

Step2: Analyze the first table (\(v_{x}\) and \(v_{y}\))

For projectile motion, the horizontal velocity \(v_{x}\) is constant (\(v_{x}=v_{0x}\)), and the vertical velocity \(v_{y}=v_{0y}-gt\). Here, \(v_{0x} = 12\ m/s\) (constant) and \(v_{y}\) changes as \(v_{y}=0 - 9.8t\) (for \(t = 1\ s\), \(v_{y}=-9.8\ m/s\); \(t = 2\ s\), \(v_{y}=-19.6\ m/s\) etc.), which is consistent with projectile motion.

Step3: Analyze the second table (\(d_{x}\) and \(d_{y}\))

For projectile motion, \(d_{x}=v_{0x}t\) (if \(v_{0x} = 8\ m/s\), \(d_{x}=8t\)) and \(d_{y}=v_{0y}t-\frac{1}{2}gt^{2}\) (here \(v_{0y} = 0\), so \(d_{y}=-\frac{1}{2}\times9.8t^{2}=-4.9t^{2}\)). For \(t = 1\ s\), \(d_{x}=8\ m\), \(d_{y}=-4.9\ m\); \(t = 2\ s\), \(d_{x}=16\ m\), \(d_{y}=-19.6\ m\) etc., which is consistent with projectile motion.

Step4: Analyze the second graph (velocity vectors)

In projectile motion, the horizontal component of velocity (\(v_{x}\)) is constant. The second graph with velocity vectors shows a non - constant horizontal component (the length of the horizontal arrow changes), which is incorrect for projectile motion (where only the vertical component of velocity changes due to acceleration \(g\) and the horizontal component remains constant).

Answer:

The second graph (the one with velocity vectors)