Question

figure a: velocity of the cart
table b (average velocity = \\( \triangle x / \triangle t \\)) for credit: show your work

Explanation:

Step1: Recall the formula for average velocity

The formula for average velocity is \(v_{avg}=\frac{\Delta x}{\Delta t}=\frac{x_f - x_i}{t_f - t_i}\).

Step2: Analyze the data for \(2^{\circ}\) track angle

From the graph, when the track angle is \(2^{\circ}\), at \(t = 0s\), \(x=0m\) (initial position \(x_i = 0m\), \(t_i=0s\)). At \(t = 1.8s\), \(x = 0.742m\) (final position \(x_f=0.742m\), \(t_f = 1.8s\)).
Substitute into the formula: \(v_{avg}=\frac{0.742 - 0}{1.8-0}=\frac{0.742}{1.8}\approx0.412m/s\)

Step3: Analyze the data for \(5^{\circ}\) track angle

Assume from the graph (since no calculation steps are shown in the table for \(5^{\circ}\) but value is given). Let's say we pick two - point: if at \(t_1\) and \(t_2\) with \(x_1\) and \(x_2\). But since the value \(v_{avg}=0.594m/s\) is provided (maybe calculated as \(\frac{x_{final}-x_{initial}}{t_{final}-t_{initial}}\) using appropriate points from the graph where \(x - t\) data is available for \(5^{\circ}\) track angle).

Step4: Analyze the data for \(10^{\circ}\) track angle

Similarly, the value \(v_{avg}=0.722m/s\) is provided (calculated using \(v_{avg}=\frac{\Delta x}{\Delta t}\) with relevant \(x - t\) data points from the \(10^{\circ}\) track - angle curve in the graph)

Answer:

For \(2^{\circ}\) track angle, the average velocity is approximately \(0.412m/s\). For \(5^{\circ}\) track angle, the average velocity is \(0.594m/s\). For \(10^{\circ}\) track angle, the average velocity is \(0.722m/s\)