Tag Archives: graphical analysis of one dimensional-motion

College Physics by Openstax Chapter 2 Problem 64


(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.

Figure 2.64

Solution:

Part A

To find the slope at t=2.5 s, we need the position values at t= 0 s and t=5 s. When t = 0 \ \text{s}, x = 0 \ \text{m}, and when t = 5 \ \text{s}, x = 17.5 \ \text{m}. The velocity at t=2.5 s is

\begin{align*}

\text{velocity} & =\text{slope} \\
\text{v} & =\frac{\Delta x}{\Delta t} \\
\text{v} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{v} & = \frac{17.5\ \text{m}-0\ \text{m}}{5 \ \text{s}-0\ \text{s}} \\
\text{v} & =3.5 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\

\end{align*}

Part B

When t = 10 \ \text{s}, x=2.5 \ \text{m}. Considering the points at t=5 s and t=10 s, the slope at 7.5 s is

\begin{align*}

\text{velocity} & =\text{slope} \\
\text{v} & =\frac{\Delta x}{\Delta t} \\
\text{v} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{v} & = \frac{2.5\ \text{m}-17.5\ \text{m}}{10 \ \text{s}-5\ \text{s}} \\
\text{v} & =-3.0 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\

\end{align*}

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College Physics by Openstax Chapter 2 Problem 63


Construct the position graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

Figure 2.18

Solution:

The figure with the corresponding examples are shown on this page: https://openstax.org/books/college-physics/pages/2-4-acceleration#import-auto-id2590556

The position-vs-time graph of the train’s motion is also graphed in the first figure here: https://openstax.org/apps/archive/20210713.205645/resources/a697c43432cdf2d09a02df47d2b746283b841fcd

(a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) The velocity of the train over time. The train’s velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.


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College Physics by Openstax Chapter 2 Problem 62


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

Figure 2.63

Solution:

To solve for the slope of the curve at t=10 s, we need two points — 1 just before and 1 just after.

When t=0 \ \text{s}, v=166 \ \text{m/s} and when t=20 \ \text{s}, v=230 \ \text{m/s}. Therefore, the acceleration is

\begin{align*}
\text{acceleration} & =\text{slope} \\
\text{acceleration} & =\frac{\Delta v}{\Delta t} \\
\text{a} & = \frac{v_2-v_1}{t_2-t_1} \\
\text{a} & = \frac{230\ \text{m/s}-166\ \text{m/s}}{20\ \text{s}-0\ \text{s}} \\
\text{a} & =3.2\ \text{m/s}^2 \\

\end{align*}

Note that the values are approximated to satisfy the given acceleration in the problem statement. The values may differ from one’s answer due to some uncertainties of a graph.


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College Physics by Openstax Chapter 2 Problem 61


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s.

Figure 2.62

Solution:

We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.

To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.

\begin{align*}

\text{slope} & =\frac{\Delta x}{\Delta t} \\
\text{slope} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{slope} & = \frac{12\ \text{m}-7\ \text{m}}{40\ \text{s}-20\ \text{s}} \\
\text{slope} & =0.25\ \text{m/s}

\end{align*}

Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertainties when using graphs. The difference is not really significant for this case.


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College Physics by Openstax Chapter 2 Problem 60


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s. Assume all values are known to 2 significant figures.

Figure 2.62

Solution:

We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.

To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.

\begin{align*}

\text{slope} & =\frac{\Delta x}{\Delta t} \\
\text{slope} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{slope} & = \frac{12\ \text{m}-7\ \text{m}}{40\ \text{s}-20\ \text{s}} \\
\text{slope} & =0.25\ \text{m/s}

\end{align*}

Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertain when using graphs. The difference is not really significant for this case.


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College Physics by Openstax Chapter 2 Problem 59


(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is 115 m/s at t=20 s. (b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is 5.0 m/s2 .

Figure 2.60
Figure 2.61

Solution:

Part A

Figure A

Figure A shows the approximate slope of the curve at time 20 seconds.

To solve for the slope of this line, we need to approximate by using two points. In this case, we shall use the points at time 15 seconds and 25 seconds.

Approximately, when t=15\ \text{s}, the position is x=1000\ \text{m}, and when t=25\ \text{s}, the position is x=2150\ \text{m}. Thefore,

\begin{align*}

\text{velocity }& = \text{slope} \\
v& = \frac{\Delta x}{\Delta t} \\
v& = \frac{x_2-x_1}{t_2-t_1} \\
v& = \frac{2150\ \text{m}-1000\ \text{m}}{25\ \text{s}-15\ \text{s}}\\
v& = 115\ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

Part B

One can immediately figure out from the given graph that it is a straight line. The slope of the line can be computed by using any two points in the line.

Here, v=15\ \text{m/s} when t=0\ \text{s}, and v=40 \ \text{m/s} when t=5\ \text{s}. The acceleration is

\begin{align*}

\text{acceleration}& = \text{slope} \\
a& = \frac{\Delta v}{\Delta t}\\
a& = \frac{v_2-v_1}{t_2-t_1} \\
a& = \frac{40\ \text{m/s}-15\ \text{m/s}}{5\ \text{s}-0\ \text{s}}\\
a& = 5.0\ \text{m/s}^2 \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

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College Physics by Openstax Chapter 2 Problem 66


Figure 2.68 shows the position graph for a particle for 6 s. (a) Draw the corresponding Velocity vs. Time graph. (b) What is the acceleration between 0 s and 2 s? (c) What happens to the acceleration at exactly 2 s?

position graph for a particle for 6 s.
Figure 2.68

Solution:

Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is composed of straight lines, we can say that the velocity is constant for several time ranges.

Time RangeSlope of the Position vs Time Graph
0 to 2 seconds=\frac{2-0}{2-0}=1\:\text{m/s}
2 to 3 seconds=\frac{-3-2}{3-2}=\frac{-5}{1}=-5\:\text{m/s}
3 to 5 seconds=0 \ \text{m/s}
5 to 6 seconds=\frac{-2-\left(-3\right)}{6-5}=\frac{1}{1}=1\:\text{m/s}

Based on the data in the table, we can draw the velocity diagram

velocity vs time graph
velocity vs time graph

Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.


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College Physics by Openstax Chapter 2 Problem 65


A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t=5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

A graph of  v(t)  is shown for a world-class track sprinter in a 100-m race.
Figure 2.67

Solution:

Part A

To find the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is

v_{\text{ave}}=6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is

a=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve.

\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by

\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec

Therefore, the total time of the sprint is

\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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