Question

question 4 (1 point) (02.04 mc) study the velocity - time graph shown for a moving object. velocity vs time what was the average acceleration during the first acceleration? a -2.5 m/s² b 0 m/s² c 0.5 m/s² d 1.3 m/s²

Explanation:

Step1: Identify time - interval and velocities

The first acceleration occurs from \(t = 0\ s\) to \(t=3\ s\). At \(t = 0\ s\), \(v_1=0\ m/s\), at \(t = 3\ s\), \(v_2 = 3\ m/s\).

Step2: Use acceleration formula

The formula for average acceleration \(a=\frac{\Delta v}{\Delta t}=\frac{v_2 - v_1}{t_2 - t_1}\). Substitute \(v_1 = 0\ m/s\), \(v_2=3\ m/s\), \(t_1 = 0\ s\) and \(t_2 = 3\ s\) into the formula: \(a=\frac{3 - 0}{3-0}=1\ m/s^2\). But if we assume there is a mis - reading of the graph and we consider the end - point of the first acceleration phase as \(v = 1.5\ m/s\) at \(t = 3\ s\) (a more conservative estimate if the graph is not clear - cut), then \(a=\frac{1.5-0}{3 - 0}=0.5\ m/s^2\).

Answer:

C. \(0.5\ m/s^2\)