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66 PE

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Found in: Page 85

### College Physics (Urone)

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.

See the step by step solution

## Understanding situation with motion graph

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.

## Calculating velocity at each time

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.