Question

1. two trials of beetle motion are shown on the graph. compare the distance, displacement, and average velocity for both trials over the course of 30 s.

motion of beetle
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Explanation:

Step1: Recall definitions

Distance is total path - length, displacement is change in position from start to end, and average velocity is displacement divided by time.

Step2: Analyze Trial 1

• Distance: Trace the curve and sum lengths. It goes up, down and then down again. Let's assume we can estimate lengths from the graph. Suppose it moves 10 cm up, 5 cm down and then 5 cm more down. So distance \(d_1=10 + 5+5=20\) cm.
• Displacement: Final position minus initial position. Initial is 0 cm and final is 0 cm. So displacement \(\Delta x_1=0\) cm.
• Average velocity \(v_1=\frac{\Delta x_1}{t}\), with \(t = 30\) s. So \(v_1=\frac{0}{30}=0\) cm/s.

Step3: Analyze Trial 2

• Distance: It moves continuously in one - direction. Suppose it reaches 25 cm at \(t = 30\) s. So distance \(d_2 = 25\) cm.
• Displacement: Initial is 0 cm and final is 25 cm. So displacement \(\Delta x_2=25\) cm.
• Average velocity \(v_2=\frac{\Delta x_2}{t}=\frac{25}{30}=\frac{5}{6}\approx0.83\) cm/s.

Step4: Compare

• Distance: \(d_2>d_1\) (25 cm > 20 cm).
• Displacement: \(\Delta x_2>\Delta x_1\) (25 cm > 0 cm).
• Average velocity: \(v_2>v_1\) (\(\frac{5}{6}\) cm/s>0 cm/s).

Answer:

For distance, Trial 2 has a greater distance than Trial 1. For displacement, Trial 2 has a greater displacement than Trial 1. For average velocity, Trial 2 has a greater average velocity than Trial 1.