In class, students worked through practice problems to prepare for the Chapter 2 test in the next class.
Students concentrated on problems 44,47,75, and 77
Monday, September 13, 2010
Saturday, September 11, 2010
Free Fall
In class we covered freely falling objects.
Ignoring frictional forces, such as air resistance, objects near the surface of the Earth will acclerate towards the Earth at a rate of -9.8 m/s^2.
Example 1:
A rock dropped from a 100 meter cliff will acclerate downwards towards the ground below.
We can calcualte the time it takes to fall in the following way.
Note in the calculation above that I used -100 meters for the displacement. Remember displacement is a vector and in free fall problem I will assign a negative sign to any downward pointing vector such as acceleration (-9.8 m/s^2) and the velocity of any downward moving object.
Example 2:
If we assume a ball thrown up from the surface of the Earth has an initial velocity of 50 m/s, it too will also experience the same acceleration as the rock in the previous problem except, in this case, the ball's positive inital velocity will continually reduced each second it rises. The ball will loose 9.8 m/s of its upward velocity each second until its velocity reaches 0 m/s, 5.1 seconds after it was initially thrown. At that instant (v=0m/s), the ball will start to fall back to the Earth gradually gaining speed as it falls.
Assuming the ball lands back where it started, the ball will impact the ground with a velocity of -50m/s at t=10.2 seconds. The time it took to travel from its apex to the ground is the same as the time it took to travel from the ground to its apex.
Homework: Read the Free Fall section:
Problems page 40: 33,35,37,39,42
Ignoring frictional forces, such as air resistance, objects near the surface of the Earth will acclerate towards the Earth at a rate of -9.8 m/s^2.
Example 1:
A rock dropped from a 100 meter cliff will acclerate downwards towards the ground below.
We can calcualte the time it takes to fall in the following way.
Note in the calculation above that I used -100 meters for the displacement. Remember displacement is a vector and in free fall problem I will assign a negative sign to any downward pointing vector such as acceleration (-9.8 m/s^2) and the velocity of any downward moving object.
Example 2:
If we assume a ball thrown up from the surface of the Earth has an initial velocity of 50 m/s, it too will also experience the same acceleration as the rock in the previous problem except, in this case, the ball's positive inital velocity will continually reduced each second it rises. The ball will loose 9.8 m/s of its upward velocity each second until its velocity reaches 0 m/s, 5.1 seconds after it was initially thrown. At that instant (v=0m/s), the ball will start to fall back to the Earth gradually gaining speed as it falls.
Assuming the ball lands back where it started, the ball will impact the ground with a velocity of -50m/s at t=10.2 seconds. The time it took to travel from its apex to the ground is the same as the time it took to travel from the ground to its apex.
Homework: Read the Free Fall section:
Problems page 40: 33,35,37,39,42
Friday, September 10, 2010
Class - 09/09/2010
Introduction to acceleration: (Sections 2.4-2.5)
Acceleration is the change in velocity of an object by either changing the objects speed or direction of motion.
On a plot describing the velocity of an object versus time, the slope of the graph, Δv/Δt, is the average acceleration of the object.
If we break this area up into two regions (see below) calculating the area of the region is simple. The blue region has an area of (vi)(Δt) while the blue area has an area of 1/2(vf-vi))(Δt).
The total area of the both the red and blue regions = (vi)(Δt) + 1/2(vf-vi))(Δt). This also represents the displacement of the object.
Δd = (vi)(Δt) + 1/2(vf-vi))(Δt).
Substituting in vi + a(Δt) for vf we get ...
Δd = (vi)(Δt) + 1/2(vi + a(Δt) - vi)(Δt).
Simplifiying, we get
Δd = (vi)(Δt) + 1/2a(Δt)^2.
Equations so far:
The last equation listed above is derived from combining other eqations and is not directly derived from the d/t graph or v/t graph. See the textbook for the derevation in section 2.5
Homework: Read 2.4-2.5
Do Problems - Page 39-40 - 21,23,25,27,and 31
Acceleration is the change in velocity of an object by either changing the objects speed or direction of motion.
On a plot describing the velocity of an object versus time, the slope of the graph, Δv/Δt, is the average acceleration of the object.
Aavg = Δv/ Δt
This equation can also be written vf = vi + a(Δt)
The diagram below is the velocity/time graph of an object that acclerates uniformly from vi to vf. The shaded area represents the displacement of the object.
If we break this area up into two regions (see below) calculating the area of the region is simple. The blue region has an area of (vi)(Δt) while the blue area has an area of 1/2(vf-vi))(Δt).
The total area of the both the red and blue regions = (vi)(Δt) + 1/2(vf-vi))(Δt). This also represents the displacement of the object.
Δd = (vi)(Δt) + 1/2(vf-vi))(Δt).
Substituting in vi + a(Δt) for vf we get ...
Δd = (vi)(Δt) + 1/2(vi + a(Δt) - vi)(Δt).
Simplifiying, we get
Δd = (vi)(Δt) + 1/2a(Δt)^2.
Equations so far:
The last equation listed above is derived from combining other eqations and is not directly derived from the d/t graph or v/t graph. See the textbook for the derevation in section 2.5
Homework: Read 2.4-2.5
Do Problems - Page 39-40 - 21,23,25,27,and 31
Class 09/10/2010
In this class I introduced the following terms: distance traveled, displacement, speed, and velocity.
I discussed the differences between distance traveled and displacement as well as the differences between speed and velocity.
1. A man walks 5 meters east.
What distance did he travel? What is his displacement?
2. A man walks 5 meters east the turns left and walks 5 meters north.
What distance did he travel? What is his displaceent? (How far is he from where he started)
3. A man walks in a circle with a 5 meters radius.
What distance did he travel? What is his displacement?
4. A man follows the path indicated in the diagram, starting at point A. The first two legs of this trip are 500m each.
Calculate the total distance traveled after each leg.
A to B
A to C
A to A
Calculate the displacement relative to A after each leg.
A to B
A to C
A to A
5. A man gets in his car and drives for 10 minutes covering a distance of 15 miles. He stops for a Snapple break for 5 minutes and then drives for 25 miles at a rate of 40 mph. Assume the man is traveling north while he is driving.
Draw a distance-time graph describing the journey.
How far did the man travel?
How long was his trip? (time)
What was his instantaneous velocity during each leg of his trip?
What was his average velocity for the trip?
6. In each of the following graphs discuss the motion of the object(s) relative to where the object began.
I discussed the differences between distance traveled and displacement as well as the differences between speed and velocity.
1. A man walks 5 meters east.
What distance did he travel? What is his displacement?
2. A man walks 5 meters east the turns left and walks 5 meters north.
What distance did he travel? What is his displaceent? (How far is he from where he started)
3. A man walks in a circle with a 5 meters radius.
What distance did he travel? What is his displacement?
4. A man follows the path indicated in the diagram, starting at point A. The first two legs of this trip are 500m each.
Calculate the total distance traveled after each leg.
A to B
A to C
A to A
Calculate the displacement relative to A after each leg.
A to B
A to C
A to A
5. A man gets in his car and drives for 10 minutes covering a distance of 15 miles. He stops for a Snapple break for 5 minutes and then drives for 25 miles at a rate of 40 mph. Assume the man is traveling north while he is driving.
Draw a distance-time graph describing the journey.
How far did the man travel?
How long was his trip? (time)
What was his instantaneous velocity during each leg of his trip?
What was his average velocity for the trip?
6. In each of the following graphs discuss the motion of the object(s) relative to where the object began.
7. Draw two lines: one representing the average velocity, the other representing the instantaneous velocity at the point marked of the graph.
Homework: Page 39 - Finish all problems 1-14.
First Day of Class 09/06/2010
In class I introduced Vavg = ∆d / ∆t
Students measured the average speed of a cart traveling at a constant velovity as well as a ball rolling down an incline.
Example data: The cart traveled 150 cm in a time of 12.5s
Example calculation: Vavg = ∆d / ∆t
Vavg = 150cm/12.5s
Vavg = 12 cm/s
Students were also asked to describe an important difference in how the two objects moved. The observation that I was hoping students would make is that the cart moved at a constat speed while the ball accleerated down the incline.
Homework: Chapter 2 - Read section 2.1-2.3
Problems 1,2,3,4,7,9a on page 39
Students measured the average speed of a cart traveling at a constant velovity as well as a ball rolling down an incline.
Example data: The cart traveled 150 cm in a time of 12.5s
Example calculation: Vavg = ∆d / ∆t
Vavg = 150cm/12.5s
Vavg = 12 cm/s
Students were also asked to describe an important difference in how the two objects moved. The observation that I was hoping students would make is that the cart moved at a constat speed while the ball accleerated down the incline.
Homework: Chapter 2 - Read section 2.1-2.3
Problems 1,2,3,4,7,9a on page 39
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