Is there anyway you can answer these questions with the information provided. Its using the moving man program.
Procedure
Place the man so that he is positioned at the 0.00 metre (m) mark. Set the initial velocity at -2.00 m/s and the acceleration at 0.50 m/s2. In a data table, record the time when the man is momentarily stopped and reversing direction.
Reset the simulation settings so that the man is once again positioned at the 0.00 m mark. Set the initial velocity at -2.00 m/s and the acceleration at 0.50 m/s2. In your data table, record the time when the man's position is at the +3.00 m mark.
Reset the simulation settings so that the man is once again positioned at the 0.00 m mark. Set the velocity at 0.00 m/s and the acceleration at 1.00 m/s2. Run the simulation and record the position of the man at 0.50 s, 1.00 s, 1.50 s, and 2.00 s.
Analysis Questions:
For step #1 of the procedure use appropriate formulae from the course to calculate the theoretical times for when the man should be momentarily stopped. Complete again for the +3.00 m mark. Compare the theoretical times with your experimental times.
Using the data collected from step #3 of the procedure, plot a graph of Displacement vs. Time. Study the shape of the graph, you can refer to the content topic of Graphical Analysis, and suggest a mathematical relationship between displacement and time if the acceleration is constant and the initial velocity is zero.
To verify that you have suggested the correct relationship in question 2, "straighten" the graph using graphical techniques. Plot the appropriate graph to verify the relationship between displacement and time when the acceleration is constant and the initial velocity is zero.
2 answers
x = xi + vi t + (1/2) a t^2
for example
"Place the man so that he is positioned at the 0.00 metre (m) mark. Set the initial velocity at -2.00 m/s and the acceleration at 0.50 m/s2. In a data table, record the time when the man is momentarily stopped and reversing direction. "
xi = 0
vi = -2
a = .5
then
v - vi + a t
v = -2 + .5 t
v = 0 at stop
so
0 = -2 + .5 t
t = 4 sec
when would he be at 3 mark?
x = xi + vi t + (1/2) a t^2
3 = 0 - 2 t + .5(.5) t^2
.25 t^2 - 2 t - 3 = 0
t^2 - 8 t - 12 = 0
t = [ 8 +/- sqrt (64 + 48) ]/2
t = [ 8 +/- sqrt (112) ]/2
t = 4 +/- sqrt 28
t = 9.3 sec