Answer: Provided the graph of distance vs time, we have to answer (a) to (f):
(a) Student's speed from 3rd to the 5th minute:
[tex]S=\frac{\Delta D}{\Delta t}=\frac{2.5-2.25}{5-3}=\frac{0.25}{2}=0.125mi/min[/tex]
(b) At what time during the bike ride did the student come to stop:
[tex]\begin{gathered} \text{ Between:} \\ 8th\rightarrow11th \end{gathered}[/tex]
(c) During what time interval was the student's speed increasing:
[tex]\begin{gathered} \text{ Between:} \\ \\ 11th\rightarrow14th \end{gathered}[/tex]
(d) What was the student's speed between 7th to 19th minute:
[tex]S=\frac{\Delta D}{\Delta t}=\frac{3.8-3.9}{19-17}=\frac{0.1}{2}=\frac{0.05mi}{min}[/tex]
(e) During what time interval was the student's speed constant:
[tex]\begin{gathered} \text{ Between the following five time intervals:} \\ 0min-1min \\ 1min-2min \\ 2min-3min \\ 3min-8min \\ 17min-23min \end{gathered}[/tex]
