By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s² at t = 10 s.

To analyze movement, motion graphs can be employed.

The mathematical approaches and graphical solutions for determining motion equations are the same.

The slope of a graph of displacement x vs. time t represents velocity v.

The inclination of a graph of velocity v vs. time t is called acceleration.

Average velocity, instantaneous velocity, and acceleration can all be calculated using graphs.

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

Coordinate (0,175) and (30,300)

Slop of the line will be

\[m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_2}}}\]

Here the coordinates are \(\left( {0,175} \right)\) and \(\left( {30,300} \right)\)

\[\begin{array}{l}m = \frac{{300 - 175}}{{30 - 0}}\\m = 4.16\,m/{s^2}\end{array}\]

Hence the acceleration is approximately to the value\[4.12 m/{s^2} \].