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66 PE

Expert-verifiedFound in: Page 85

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Figure 2.68 shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graph **

As the velocity is constant in the interval t = 0 to 6 s, the acceleration for this motion is zero.

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 constant during \[t{\rm{ }} = 0\] to \(6\) s, the acceleration for this** **motion is zero**.**

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