course Phy 201
QUIZAn object travels a distance of 32 cm in 17.6 seconds from start to finish. If its acceleration is uniform and its initial velocity zero, then what are its average velocity, acceleration and final velocity?
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D = 32 cm
T = 17.6 s
V initial = 0
vAve= change in x / change in t = 32 cm/17.6 sec = 1.8 cm/sec
acceleration = change in velocity / time elapsed
s = at^2/2
32 cm = a 309.76 /2
64 cm = 619.52 a
acceleration = 0.103 cm/sec^2
final velocity = (acceleration)(Time)
final velocity = (0.103 cm/sec^2 ) (17.6 sec)
final velocity = 1.81 cm /sec
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Except for an error in your algebra you would have obtained correct solutions. However you used some 'canned' formulas that do apply to this situation, but which can easily be misused, and you did not follow a reasoning process that will lead to a deeper understanding of these quantities.
Since acceleration is uniform and initial vel is 0, the final vel is double the initial vel. You should understand why this is so, in terms of the v vs. t graph. So vf = 3.6 cm/s.
Acceleration is rate of change of velocity with respect to clock time, which is therefore (3.6 cm/s - 0 cm/s) / 17.6 sec = 0.21 cm/s^2, approx.
course Phy 201
This is the first quiz. I named the Problem #3 as Quiz #1 previously by accident.Also, I sketched a graph in paint and added it to my word document, however when I copied and pasted it did not trasfer to the submit work form.
QUIZ
If an object increases velocity at a uniform rate from 5 m/s to 21 m/s in 12 seconds, what is its acceleration and how far does it travel?
Sketch a velocity vs. clock time graph for an object whose initial velocity is 5 m/s and whose velocity 12 seconds later is 21 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
Change in velocity / change in time
21 m/s – 5 m/s / 12 s = 16 m/s / 12 s = 1.33 m/s^2
This graph demonstrates the problem above.
The x-axis is TIME (seconds), The y-axis is VELOCITY (m/s)
The coordinates for this graph were 12, 21 and 0,5
To get the slope I subtracted 12 – 0 = 12 (rise) 21 – 5 = 16(run)
Slope = 12/16 = 0.75
The is a positive slope, and the graph is increasing.
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Good explanation of the slope. What does this quantitiy 1.33 m/s^2 mean in terms of the motion of the object?
You didn't explain the area. Please submit a revision answering my question above and explaining the calculation and the meaning of the area.