91Ó°ÊÓ

The position versus time graph is

For a position vs time graph, a slope of the line that is tangent to the graph at time t is the instantaneous velocity at that time.

For $t=10s$,

consider points, $\left(t_1,x_1\right)=\left(0,25\right)$and $\left(t_2,x_2\right)=\left(20,50\right)$

The slope of the graph gives the velocity of the jogger at that point.

localid="1648057941837" $$\begin{gathered} v=\frac{x_2-x_1}{t_2-t_1} \\ =\frac{50-25}{20-0} \\ =\frac{25}{20} \\ =1.25m/s \end{gathered}$$

Therefore, jogger's velocity at $t=10s$is $1.25m/s$.

For $t=25s$,

consider points, $\left(t_1,x_1\right)=\left(20,50\right)$and $\left(t_2,x_2\right)=\left(30,50\right)$

The slope of the graph gives the velocity of the jogger at that point.

$$\begin{gathered} v=\frac{x_2-x_1}{t_2-t_1} \\ =\frac{50-50}{30-20} \\ =0m/s \end{gathered}$$

Therefore, the jogger's velocity at$t=25s$is $0m/s.$

For $t=35s$,

consider points$\left(t_1,x_1\right)=\left(30,50\right)$and$\left(t_2,x_2\right)=\left(40,0\right)$

The slope of the graph gives the velocity of the jogger at that point.

$$\begin{gathered} v=\frac{x_2-x_1}{t_2-t_1} \\ =\frac{0-50}{40-30} \\ =-5m/s \end{gathered}$$

Therefore, the jogger's velocity at $t=35s$is$-5m/s$.