We are given a table of velocity vs time. We are asked the following:
Part a. We are asked to determine the average acceleration. To do that we will use the following formula:
[tex]a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_0}{t_f-t_0}[/tex]
Where:
[tex]\begin{gathered} a=\text{ acceleration} \\ v_f,v_0=\text{ final and initial velocities} \\ t_f,t_0=\text{ final and initial time} \end{gathered}[/tex]
Now, we use the final and initial values of velocity and time from the table.:
[tex]a=\frac{1.14\frac{mm}{ms}-0.2\frac{mm}{ms}}{650ms-50ms}[/tex]
Solving the operations:
[tex]a=0.0016\frac{mm}{ms^2}[/tex]
Therefore, the acceleration is 0.0016 mm/ms^2.
To convert to m/s^2 we will use the following conversion factors:
[tex]\begin{gathered} 1000mm=1m \\ 1000ms=1s \end{gathered}[/tex]
Now, we multiply by the conversion factors:
[tex]a=0.0016\frac{mm}{ms^2}\times\frac{1m}{1000mm}\times(\frac{1000ms}{1s})^2[/tex]
Solving the operations;
[tex]a=1.6\frac{m}{s^2}^{}[/tex]
Part B. We are asked to determine the displacement. The displacement in a graph of velocity vs time is equivalent to the area under the curve of the graph, like this:
The area under the curve is approximately the area of a triangle, therefore, its area is given by:
[tex]d=\frac{bh}{2}[/tex]
Where the base is the difference in time and the height is the difference is velocity, therefore, we have:
[tex]d=\frac{\Delta t\Delta v}{2}=\frac{(t_f-t_0)(v_f-v_0)}{2}[/tex]
Substituting the values we get
[tex]d=\frac{(650ms-50ms)(1.14\frac{mm}{ms}-0.2\frac{mm}{ms})}{2}[/tex]
Solving the operations:
[tex]d=282mm[/tex]
Now, we convert mm to cm using the following conversion factor:
[tex]10mm=1cm[/tex]
Multiplying by the conversion factor we get:
[tex]d=282mm\times\frac{1cm}{10mm}=28.2cm[/tex]
Therefore, the displacement is 28.2 cm.
Part C. This is a uniformly accelerated motion or a motion with a constant acceleration. We can tell from the graph because in this type of motion the graph of velocity vs time is a straight line. The slope of the line is the acceleration of the motion due to the fact that the acceleration does not change with time.
