Tag Archives: position vs time graph

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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