class 050926
Text Homework:
Phy 121: Do All Level I
problems in Chapter 2.
Phy 201: Do the following problems
in Chapter 2: 6, 9, 11, 16, 22, 27, 37, 43, 52, 55, 58, 63
note access links to questions:
18-34-375 (Principles of Physics, General College Physics,
University Physics)
11-09-453 (General College Physics, University Physics)
23-24-380 (University Physics only)
Write down the four equations of uniformly accelerated motion
and show how you check the units for each.
If a marble is given an initial velocity on a long uniform
incline on which the magnitude of its acceleration is 40 cm/s^2, and if it passes the point 20 cm
up the ramp 2 seconds later, then
- Which of the quantities v0, vf, a, `dt and `ds do we
know?
- We know a = 40 cm/s^2 or -40 cm/s^2, `ds = 20 cm and
`dt = 2 s.
- What equation would we use to find a fourth quantity?
- The 3d contains a, `ds and `dt so it will give us a
fourth quantity v0.
- What do you get if you solve this equation for that
fourth quantity?
- Solving the 3d equation for v0 we get v0 = `ds / `dt -
1/2 a `dt.
- What do you get when you substitute the values of the
three known quantities?
- We get v0 = -30 cm/s.
- What then is (are) the value (s) of the fifth quantity?
- We get vf = 50 cm/s, using direct reasoning with flow
diagrams.
- We get the same thing using the equations of motion.
- What does all this tell you about the motion of the
marble on the incline?
Determine the acceleration of an object whose velocity is initially 19cm/s and which accelerates uniformly through a distance of 75cm in 5.8 seconds.
Would I just find the velocity for 75cm and 5.8 seconds and subtract that from the other velocity or what? Or would i just take 19cm/s/5.8s and get 3.3cm/s^2 and then take 78cm/5.8s and then take the velocity and divide that by 5.8s again to get 2.3cm/s^2
Of the five quantities v0, vf, `dt, `ds and a which are you given the values of?
Which of the four equation(s) of uniformly accelerated motion apply?
What results do you get?
An object is given an unknown initial velocity up a long ramp on which its acceleration is known to have magnitude 6cm/s^2 .223 seconds later it passes for the first time, a point 5cm up the ramp from its initial position.
Of the five quantities v0, vf, `dt, `ds and a which are you given the values of?
Which of the four equation(s) of uniformly accelerated motion apply?
What results do you get?
I am really just completely lost on this problem, could you show me how to work this one!? THanks
What is its initial velocity and how much longer will it be before it passes this point again?
Am I supposed to be using one of the 4 equations you gave us?
This is solved very much like the problem I solved in class on Friday. Identify your variables and solve the equations.
major quiz question __ is accel independent of position and velocity
If a cart coasts distances of 8.22996, .9279166, 7.202417, and 6.310089cm,
starting from rest each time, and requires respective times of 9.5sec, 1sec,
8.25sec, and 6.75sec, is the hypothesis that acceleration is independent of
position or velocity supported or not?
So i'd find the velocities & accel 8.29cm/9.5sec= .87cm/sec .09cm/s^2
9.3cm/1sec= 9.3cm/sec 9.3cm/s^2
that's .93 cm, not 9.3 cm
7.2cm/8.25s= .87cm/sec .87cm/s^2
the average acceleration will not be equal to the average velocity. Was this
a typo?
6.3cm/6.75s= .93cm/sec .14cm/s^2
Pretty good, but it looks like you divided average velocity by change in
clock time to get acceleration, which is not correct.
Average acceleration is change in velocity / change in clock time, which is not
at all the same thing as average velocity / change in clock time.
Now I don't
know how to say these relate, the accelerate relies on the position and the
velocity, according the velocity is change in position/change in time.
Are there significant differences in the accelerations? If so then
presumably these differences would be due to either positions or velocities.
How far does an object travel and what is its acceleration if its velocity
increases at a uniform rate from 7m/s to 25m/s in 10 seconds?
velocity = 7m/s to 25m/s in 10 sec
25m/s-7m/s * 10s
You multiply average velocity by change in clock time, not change in
velocity by change in clock time.
You have two velocities. How do you get the average velocity?
18m/s * 10s=180m
Explain the meaning of the slope and area of the velocity vs. time for this
object during this time interval?
What would be the coordinates of the two points on the v vs. t graph?
Sketch the graph and sketch the trapezoid defined by these two points.
What is the slope of the line segment between the points and what does it
mean?
What is the area of the trapezoid and what does it mean?
Slope and area are
the same thing. So is that what you're asking for?
Slope and area are unrelated. They aren't the same thing, and they have very
different meanings.
My question is Ch.2 #27 -- A car traveling 85 km/h
strikes a tree. The front end of the car compresses and the driver comes to rest
after traveling 0.80 m. What was the average acceleration of the driver during
the collision? Express the answer in terms of g's where 1.00g = 9.80 m/s/s. The
first thing I did was change 85 km/h to 23.6 m/s, then I decided that for the
driver 'ds=.80m and vf=0m/s. I am not sure where to go from here. If we know 'ds
and vf of the driver we need vo to solve for a. I am not sure how to interpret
the 85 km/h(23.6 m/s) speed of the car.
You are correct that you need v0. The problem gives you v0:
You have ds and vf for the .80 m displacement. At the instant the car first
contacts the tree it is moving at 85 km/hr, or 23.6 m/s. So v0 = 23.6 m/s.
Thus you have the three variables you need to calculate the acceleration.