Question

2.Given the velocity versus time graph shown below, sketch the corresponding position versustimeand acceleration versus timegraphs. (Assume the object starts at x= 0 m at time t= 0 s).Be sure to label bothaxes of each graph with the correctscale.
3.Given the acceleration versus timegraph shown below, sketch the corresponding velocity versus timeand position versus timegraphs. Assume that initial velocity and position at t= 0 s is equal to 0 m/s and 0 m respectively. Be sure to label bothaxes of each graph with the correctscale.

Answer 1

Graphical analysis of motion gives the graphical the relationships between position, velocity, and acceleration, versus time

2. Please find attached the required position versus time and acceleration versus time graphs

3. Please find attached the required velocity versus time and position versus time graphs

The reasons the attached graphs are correct are given as follows:

2. The given coordinates of the vertex points obtainable from the graph is first written as follows:

$\begin{array}{|c|c|c|c|c|}\underline {Time, \ t}&\underline {Velocity, \ v}&Acceleration, a&\underline {Distance , \ s =v\cdot \Delta t+\dfrac{1}{2} \cdot a \cdot (\Delta t)^2}&Position\\0&-1&0&0&0\\2&1&1&0&0\\4&-1&-1&0&0\\6&1&1&0&0\\10&1&0&4&4\\12&0&-0.5&1&5\end{array}$The distance covered between time intervals of time, s, is given as follows;

$s = v \cdot \Delta t + (1/2) \cdot a \cdot ( \Delta t)^2$

The position, Pₙ = s₁ + s₂ + s₃ + ... + sₙ

Other values on the graph obtained by calculation on a spreadsheet are;

$\begin{array}{|l|cl|}Time \ (s)&&Position \ (m)\\0.5&&-0.375\\1&&-0.5\\1.5&&-0.375\\2.5&&0.375\\3&&0.5\\3.5&&0.375\\4.5&&-0.375\\5&&-0.5\\5.5&&-0.375\\6.5&&0.5\\7&&1\\7.5&&1.5\\8&&2\\8.5&&2.5\\9&&3\\10&&4\\10.5&&4.4375\\11&&4.75\\11.5&&4.9375\\12&&5\end{array}\right]$

Please find attached the acceleration versus time graph

3. The coordinates of the points are presented as follows:

$\begin{array}{|c|cc|}\underline{Time\ (s)}&&\underline{Acceleration\ (m/s^2)}\\0&&0.5\\2&&0.5\\2&&0\\4&&0\\4&&-0.5\\6&&-0.5\\6&&0\\8&&0\\8&&0.5\\10&&0.5\\10&&-0.5\\12&&-0.5\end{array}\right]$

The distance covered between time intervals, s, is given as follows;

$s = v \cdot \Delta t + (1/2) \cdot a \cdot ( \Delta t)^2$

The position, Pₙ = s₁ + s₂ + s₃ + ... + sₙ

Using a spreadsheet application, more detailed values of the position can be found as shown in the graph created with MS Excel

• Time

0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6, 6.5, 7, 7.5, 8, 8.5, 9, 9.5, 10, 10.5, 11, 11.5, 12

• Position

0, 0.06, 0.25, 0.56, 1.00, 1.50, 2, 2.5, 3, 3.5, 3.81, 4.00, 4.06, 4.06, 4.06, 4.06, 4.06, 4.13, 4.31, 4.63, 5.06, 5.50, 5.81, 6, 6.06

Learn more about graphing motion here:

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