class 050921
Testing:
Briefly, go to the Learning Lab, go to a computer, your homepage at http://vhmthphy.vhcc.edu/ and click on Tests.
Choose your course and click on Major Quiz.
Print off the test.
Sign the danged thing.
Get it signed by the attendant.
Take it in the room to which the attendant directs you.
Hand it to the attendant.
Experiment:
Set up a ramp system to project a marble off the edge of a table in the horizontal direction and with a constant horizontal velocity.
Observe the horizontal range as the marble falls to the floor.
Observe the horizontal range over a much shorter fall.
Calculate average horizontal velocities for both falls (determine time of fall for each trial and divide time of fall into horizontal range, as shown in Intro Prob Sets and in today's notes).
Write down the equations of uniformly accelerated motion (2 minutes).
Most of the class is doing well with this question, but if you are not you need to contact me and explain what the difficulty is.
Be sure you get the equations completely right. Be sure that the units always check out.
Sketch a graph of v vs. t for an object which travels down a 50 cm incline, starting from rest, in 5 seconds. Find the slope and area of the graph.
Ave vel. is
ave rate of change of position with respect to clock time, which is
change in position / change in clock time or
`ds / `dt.
In this example ave. vel. is 50 cm / (5 s) = 10 cm/s.
The v vs. t graph is a straight line because accel is uniform.
Since v0 = 0 (starting from rest) the graph will start at the origin (v = 0 at t = 0).
The average value of the velocity is 10 cm/s, so the straight line goes up into the first quadrant.
The motion ends at t = 5 sec so the line extends out to t = 5 sec.
Since the graph is a straight line, and since the average vertical coordinate of a v vs. t graph represents average velocity, the average velocity occurs at the midpoint of the line segment.
Since init vel is 0 and the midpoint vel. is 10 cm/s, the final vel. is 20 cm/s.
The slope is therefore rise / run = 20 cm/s / (5 s) = 4 cm/s^2.
What does slope represent?
Rise represents change in velocity.
Run represents change in clock time.
So rise / run represent change in velocity / change in clock time.
change in vel / change in clock time is ave rate of change of vel with respect to clock time.
ave rate of change of vel with respect to clock time is acceleration, by the definition of acceleration.
The area is ave. 'altitude' * width, which is 10 cm/s * 5 s = 50 cm.
What does the area represent?
Area is ave. 'altitude' * width.
Altitudes of a v vs. t graph represent velocities so ave 'altitude' represents ave. vel.
Width of a v vs. t trapezoid represents change in clock time.
So ave altitude * width represents ave vel. * change in clock time.
Since ave. vel. is ave rate of change of position with respect to clock time, ave. vel. = change in position / change in clock time.
So ave. vel. * change in clock time = (change in position / change in clock time ) * change in clock time = change in position.
What would be the units of each of the following: