a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t = 2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Motion graphs can be used to analyze movement.

For determining motion equations, the mathematical procedures and graphical solutions are the same.

The velocity v is represented by the slope of a graph of displacement x vs. time t.

Acceleration is the inclination of a graph of velocity v vs. time t.

Graphs can be used to calculate average velocity, instantaneous velocity, and acceleration.

Here if we carefully look into the figure, in your text book lets take any two points on the line

Coordinate \(\left( {0,0} \right)\) and \(\left( {7,25} \right)\)

Slop of the line will be

\[\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(0,0)\,\,\,and\,\,\,(7,25)\\m = \frac{{25 - 0}}{{7 - 0}}\\m = 3.57\,m/s\end{array}\]

The velocity of the graph at 2.5 time is \[3.57 m/s\] approx.

Here if we carefully look into the figure, in your text book lets take any two points on the line

Coordinate \(\left( {11,0} \right)\) and \(\left( {4,25} \right)\)

Slop of the line will be

\[\begin{array}{l}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\\Here\,the\,coordinates\,are\,(11,0)\,\,\,and\,\,\,(4,25)\\m = \frac{{25 - 0}}{{4 - 11}}\\m = - 3.57\,\,m/s\end{array}\]

The velocity of the graph at 7.5 s time is \[3.57 m/s\] approx.

Hence the value which we derived also consist the figure 2.64.