63 PE - College Physics (Urone)

Question

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

Step-by-Step Solution

Answer

The graph obtained will be non linear 

Creating graph and its analysis

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 t = 1, 2, 3, 4, 5.......s

Here if we carefully look into the figure, the average acceleration is around 1.8 m/s²

\(\begin{array}{l}{v_1} = a{t_1}\\{v_1} = 1.8 \times 1\\{v_1} = 1.8\,m/s\end{array}\)

Hence the velocity at 1s is 1.8m/s

Similarly calculating for \(t = {\rm{ }}2{\rm{ }}s\) 

\(\begin{array}{l}{v_2} = a{t_2}\\{v_2} = 1.8 \times 2\\{v_2} = 3.6\,m/s\end{array}\)

Hence the velocity at 2 s is 3.6 m/s

\(\begin{array}{l}{v_3} = a{t_3}\\{v_3} = 1.8 \times 3\\{v_3} = 5.4\,m/s\end{array}\)

Hence the velocity at 3 s is 5.4 m/s.

\(\begin{array}{l}{v_4} = a{t_4}\\{v_4} = 1.8 \times 4\\{v_4} = 7.2\,m/s\end{array}\)

Hence the velocity at 4 s is 7.2 m/s.

\(\begin{array}{l}{v_5} = a{t_5}\\{v_5} = 1.8 \times 5\\{v_5} = 9\,m/s\end{array}\)

Hence the velocity at 1 s is 1.8 m/s.

Calculating the distance at t = 1, 2, 3, 4, 5….s

The displacement can be calculated using the formula

\(d = ut + \frac{1}{2}a{t^2}\)

Initial velocity =0

\(\begin{array}{l}{d_1} = \frac{1}{2}a{t_1}^2\\{d_1} = \frac{1}{2}\left( {1.8} \right){\left( 1 \right)^2}\\{d_1} = 0.9\,m\\{d_1} = 0.0009\,km\end{array}\)

The displacement in 1 s is 0.0009 km

\(d = ut + \frac{1}{2}a{t^2}\)

Initial velocity = 0

\(\begin{array}{l}{d_2} = \frac{1}{2}a{t_2}^2\\{d_2} = \frac{1}{2}\left( {1.8} \right){\left( 2 \right)^2}\\{d_2} = 3.6\,m\\{d_2} = 0.0036\,km\end{array}\)

The displacement in 2 s is 0.0036km

\(d = ut + \frac{1}{2}a{t^2}\)

Initial velocity = 0

\(\begin{array}{l}{d_3} = \frac{1}{2}a{t_3}^2\\{d_3} = \frac{1}{2}\left( {1.8} \right){\left( 3 \right)^2}\\{d_3} = 8.1\,m\\{d_3} = 0.0081\,km\end{array}\)

The displacement in 3 s is 0.0081km

\(d = ut + \frac{1}{2}a{t^2}\)

Initial velocity = 0

\(\begin{array}{l}{d_4} = \frac{1}{2}a{t_4}^2\\{d_4} = \frac{1}{2}\left( {1.8} \right){\left( 4 \right)^2}\\{d_4} = 14.4\,m\\{d_4} = 0.0144km\end{array}\)

The displacement in 4 s is 0.0144km

Now let’s plot it in the graph

Hence the above graph represents the answer of the question.