Question

mathematics: show the calculations for one set of data above.
v_a=\frac{1.9 cm}{t_a}
v_b=\frac{1.9 cm}{t_b}
a=\frac{v_b - v_a}{t_{ab}}

Explanation:

Step1: Assume values for \(t_A\) and \(t_B\)

Let \(t_A = t_B=t\) (if time is the same for both cases as not otherwise specified). Then \(v_A=\frac{1.9\ cm}{t}\) and \(v_B = \frac{1.9\ cm}{t}\), so \(v_A = v_B\).

Step2: Calculate acceleration \(a\)

Since \(v_A = v_B\), then \(a=\frac{v_B - v_A}{t_{AB}}=\frac{0}{t_{AB}} = 0\) (assuming \(t_{AB}\) is non - zero).

Answer:

If \(t_A=t_B\), the acceleration \(a = 0\). If \(t_A
eq t_B\), we need specific values of \(t_A\), \(t_B\) and \(t_{AB}\) to calculate \(v_A\), \(v_B\) and \(a\) accurately. For example, if \(t_A = 1\ s\) and \(t_B=2\ s\):

Step1: Calculate \(v_A\) and \(v_B\)

\(v_A=\frac{1.9\ cm}{1\ s}=1.9\ cm/s\), \(v_B=\frac{1.9\ cm}{2\ s}=0.95\ cm/s\)

Step2: Calculate acceleration \(a\)

Let \(t_{AB}=1\ s\), then \(a=\frac{v_B - v_A}{t_{AB}}=\frac{0.95 - 1.9}{1}=- 0.95\ cm/s^{2}\)