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graph velocity b (v_b) vs time a to b (t_ab) on the back side of this p…

Question

graph velocity b (v_b) vs time a to b (t_ab) on the back side of this paper. make a graph of velocity versus time. velocity should be on the vertical axis (v). time should be on the horizontal axis (t_ab). analyze: describe the line or 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_{ab}=\frac{v_b - v_a}{t_{ab}}

Explanation:

Step1: Identify variables

We have velocities $v_A=\frac{1.9\ cm}{t_A}$ and $v_B = \frac{1.9\ cm}{t_B}$, and acceleration $a=\frac{v_B - v_A}{t_{AB}}$. But we need to assume some values for time to calculate. Let's assume $t_A = 1\ s$ and $t_B= 0.5\ s$.

Step2: Calculate velocities

$v_A=\frac{1.9\ cm}{1\ s}=1.9\ cm/s$ and $v_B=\frac{1.9\ cm}{0.5\ s} = 3.8\ cm/s$.

Step3: Calculate acceleration

$t_{AB}=t_B - t_A=0.5 - 1=- 0.5\ s$ (assuming $A$ occurs before $B$). Then $a=\frac{v_B - v_A}{t_{AB}}=\frac{3.8 - 1.9}{-0.5}=\frac{1.9}{- 0.5}=-3.8\ cm/s^{2}$.

Answer:

If we assume $t_A = 1\ s$ and $t_B = 0.5\ s$, then $v_A = 1.9\ cm/s$, $v_B=3.8\ cm/s$ and $a=-3.8\ cm/s^{2}$