91Ó°ÊÓ

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 we can see in the graph that there is a trend of increasing velocity in time \[t = 0\] to \[t = {\rm{ }}2s\].

For this, the time velocity is constant; let’s take coordinates \(\left( {0,0} \right)\) and \(\left( {2,2} \right)\).

\(\begin{array}{l}m = \frac{{{Y_2} - {Y_1}}}{{{x_2} - {x_1}}}\\m = \frac{{2 - 0}}{{2 - 0}}\\m = 1\,m/s\end{array}\)

Hence the velocity at \(0\) to \(2{\rm{ s}}\)is \[1{\rm{ }}m/s\].

Similarly, here we can see in the graph that there is a trend of decreasing trend velocity in time \[t = 2\] to \[t = {\rm{ }}3s\].

For this time, velocity is constant; let’s take coordinates \(\left( {2,2} \right)\) and \(\left( {3, - 3} \right)\)

\(\begin{array}{l}m = \frac{{{Y_2} - {Y_1}}}{{{x_2} - {x_1}}}\\m = \frac{{ - 3 - 2}}{{3 - 2}}\\m = - 5\,\,m/s\end{array}\)

Hence the velocity for the interval \(2\) to \(3 s\) is \[ - 5{\rm{ }}m/s\].

Similarly, here we can see in the graph that there is a trend of constant velocity in time interval \(t = 3\) to \(t = 5 s\).

For this time, velocity is constant; let’s take coordinates \(\left( {3, - 3} \right)\) and \(\left( {5, - 3} \right)\).

\(\begin{array}{l}m = \frac{{{Y_2} - {Y_1}}}{{{x_2} - {x_1}}}\\m = \frac{{ - 3 + 3}}{{5 - 3}}\\m = 0\,\,m/s\end{array}\)

Hence the velocity during t = \(3\) to \(5 s\) is \[0{\rm{ }}m/s\].

Similarly, here we can see in the graph that there is a trend of increasing velocity in duration \(t = 5 s\) to \(t = 6 s\).

For this time, velocity is constant; let’s take coordinates \(\left( {6, - 2} \right)\) and \(\left( {5, - 3} \right)\).

\(\begin{array}{l}m = \frac{{{Y_2} - {Y_1}}}{{{x_2} - {x_1}}}\\m = \frac{{ - 3 + 2}}{{5 - 6}}\\m = 1\,\,m/s\end{array}\)

Hence the velocity at the interval \(5\)to \(6 s\) is \[1{\rm{ }}m/s\].

Hence the above figure shows the velocity of the system.

As the velocity is constantduring \[t{\rm{ }} = 0\] to \(6\) s, the acceleration for thismotion is zero.