course PHY 121
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What are the average acceleration and final velocity of an object which accelerates uniformly from rest, attaining a velocity of 40 cm/sec in 13.6 seconds? How far does the object travel during the 13.6 seconds?
Since the objec accelerates from rest, the initial velocity is 0 cm/s. The final velocity is the velocity at the end of the time interval which is 13.6 seconds. The final velocity was given to us, 40 cm/sec. Acceleration is the average rate of change in velocity. It is expressed by the formula: `dv/`dt. So the average acceleration is (40 - 0)/13.6 = (40cm/s)/ 13.6s = 2.94 cm/s/s = 2.94 cm/s^2. The distance traveled can be found by multiplying the time interval by the change in velocity: (13.6s)(40cm/s)= 544 centimeters.
Multiplying time interval by change in velocity does not give a meaningful quantity.
Since vAve = `ds / `dt, the we see that `ds = vAve * `dt: displacement = average velocity * time interval.
Please submit a revised solution to the last part of this problem, and be sure to include my notes.
WEEK 2, QUIZ 1:
If an object increases velocity at a uniform rate from 7 m/s to 21 m/s in 11 seconds, what is its acceleration and how far does it travel?
Acceleration is the average rate of change in velocity. It is expressed by the formula: `dv/`dt. The change in velocity is: 21m/s - 7m/s = 14m/s. The change in time is 11 seconds.
Acceleration = (14m/s)/11s = 1.27m/s^2.
Distance traveled can by found by multiplying time interval, 11 seconds, by the change in velocity: (11s)(14m/s) = 154 cm
Sketch a velocity vs. clock time graph for an object whose initial velocity is 7 m/s and whose velocity 11 seconds later is 21 m/s. Explain what the slope of the graph means and why, and also what the area means and why.
**I have emailed my graph.
The slope of the velocity vs. clock time represents the rate of change in velocity or acceleration. The area under the graph represents average velocity times time interval which is the change in position. The average altitudes of the intervals or trapezoids under the graph represents average velocity and the width represent the time interval.
You need to be very specific in your description of the meaning of your graph. Everything you say here is correct, but you should specify the exact rise and run of your graph, describe exactly how they give you the slope and what the result is, etc.. Similar detail should be included in your interpretation of the area.
Please submit that information along with the other details of this problem, including my notes.
Most of your answers are good, and you are very close to putting all this together very nicely. Please resubmit the requested work and let me know if you have questions.