Displacement: the distance that an object travels with respect to its origin.

Velocity: vector measurement of the how far an object is displaced over a period of time

Acceleration: change in velocity (can be in speed OR direction!)

Change in distance in the presence/absence of acceleration

Without acceleration

Constant velocity --> Constant change in distance

With Acceleration

Distance changes exponentially while velocity changes at a constant rate

(Think of the graphs: slope of D v. T graph is velocity; slope of V v. T graph is acceleration. Acceleration makes exponential D v. T graph)

+/- values for each of these

Subjective references of direction

Displacement:

+: Object moved closer to/away from the origin

-: Object moved in a direction that is opposite to what is defined as the positive direction

Velocity

+: Object moving in positively defined direction

-: Object moving in negatively defined direction

Acceleration

+: Velocity changes to become more positive

-: Velocity changes to become more negative

Acceleration and Velocity in same direction --> speed increases

Acceleration and Velocity in different direction --> speed decreases

Graphs versus Time

Displacement

0 slope: Object stays at rest for period of time

+ slope: Object moves from origin in positively defined direction

- slope: Object moves from origin in negatively defined direction

Denotes constant velocity of slope is constant

Denotes acceleration if graph is exponential function

Velocity

0 slope: Object moves with constant velocity

+/- slope: Object accelerates

Denotes constant acceleration if slope of a line is constant

Object increases speed if the line moves away from the X-axis

Object decreases speed if the line moves toward the X-axis

Acceleration

0 slope: Object moves with constant acceleration

+/- slope: Object moves with changing acceleration

Acceleration and Velocity in same direction --> speed increases

Acceleration and Velocity in different direction --> speed decreases

Relative Motion

Put simply, it relies on the use of reference points

If the motion of anything confuses you, just reimagine the situation on a train (usually works out)

Scale Drawings

Be sure to be consistent with scale

35 degrees E of N notation (move 35 degrees east from North)

Vectors/trig

Seperation of velocity/distance into components along the x and y axis with a given angle from the horizontal

Use of various trig functions to find mssing components

Major Equations/Constants:
V= d/t (also d =vt or t=d/v)
d=Vit+1/2at^2
Vf=Vi+at
(Vf)^2=(Vi)^2+2ad

d = Distance

v= Velocity

Vi = Initial Velocity

Vf = Final Veloctiy

a = Acceleration

t = Time

Acceleration due to gravity = 9.8 m/s/s

Investigation:

An investigation that we did for this unit was the "Shoot Your Grade" experiment. This investigation caused our class to apply our knowledge of kinematics and projectile motion to predict the displacement of a marble shot from a catapult. First, each group was given a specific angle and setting for power on the catapult. The groups then recorded the time it took for the marble to be fired from the catapult, and using the diameter of the marble as "d" (I forget what it was), the initial velocity of the marble was found by dividing by the newly found time, "t". Using components of a right triangle, one could use this velocity to be the hypotenuse of a right triangle, shot from x degrees from the horizontal. However, since the angle was initially given to each group, one could use a trig function to find the velocity of the X and Y components (v cos x for horizontal, v sin x for vertical).

After all the basic information is found, there are a number of ways to approach this problem. The way that I did this was by first calculating the total amount of time that the marble was airborne and by multiplying this time by the horizontal velocity as calculated from the components. This would work assuming that there are ideal conditions (No air resistance, meaning that there would be no force in the X-direction acting on the marble. The marble, therefore, would travel with a constant velocity in the X direction.). To find the time, one would need to start by finding the time it took the marble to reach its maximum height. Then, by using the maximum height, the remaining time would be would by calculating how much time it would take for the marble to fall to the ground from it's maximum height. The velocity in the X component would already be calculated.

The first step one would take would be to find the maximum height of the marble. Since the initial velocity in the Y (vertical) direction was already found through the use of components, and since we know the final velocity in this direction is 0 m/s/s and acceleration the ball would face due to gravity (9.8 m/s/s), the time it would take the marble to reach this height can be calculated by using the formula Vf=Vi+at, which would be rearranged to be t = (Vf-Vi)/a. An important thing to note is that the acceleration due to gravity should be denoted with a negative sign to show that it is against the current motion of the projectile. This opposition should be shown either in this way (which is probably easier) or by changing to initial velocity to a negative number. Once the time is found, there is enough information to find the height by using the formula d=Vit+1/2at^2, where d is the height (vertical distance, technically). Again, there should be representation of the acceleration due to gravity opposing the initial motion of the marble by using a negative sign. Once this value is found, the last thing to do to measure the height is to measure the distance of the catapult from the floor and add this value to the calculated value (since the calculations show the distance from the catapult, not the floor).

After finding this height, one would need to calculate the time it takes for the marble to fall to the ground. The initial velocity would be zero since, in the Y component, there is no velocity, (though the marble is still moving horizontally). The height and acceleration due to gravity are already known, so the equation (Vf)^2=(Vi)^2+2ad can be used to find the velocity of the marble right before it hits the ground (to get Vf by itself, though, you would need to take the square root of both sides of the equation first). With this velocity, the equation Vf=Vi+at can be rearranged to find the time it takes for the marble to acquire this speed [t = (Vf-Vi)/a]. The next thing to do is to add this time with the previously found time it takes for the marble to reach its maximum height to get the total amount of time that the marble was airborne.

With the total time the marble is in the air and the unchanging velocity of the marble in the X component, one can use the equation d=vt to calculate the horizontal distance at which the marble would hit the ground. As stated before, this equation assumes ideal conditions. To account for air resistance, one could make a rough estimate by subtracting between 5% and 10% of the calculated distance.

Open Response Question
A ball rests on top of an inclined plane. It is given a slight push and rolls down the plane and onto a frictionless table top. The ball then rolls off the table and falls onto the ground.
A. At each stage of the marble's motion, state whether there is a constant acceleration, a constant velocity, or neither. Explain your reasoning for each.

At the top of the inclined plane

Rolling down the plane

Rolling across the table

Falling off the table

B. The ball accelerates a.75 m/s/s down the ramp, beginning at the table with a speed of 6 m/s. How long is the ramp?

C. How long did it take the ball to attain this velocity?

D. What would the velocity be if the ball were initially pushed to go 2 m/s down the ramp? How long would it take the ball to reach 6 m/s?

E. The ball rolls off the table and is airborne for 2.5 seconds before it hits the ground. Which calculated velocity of the tabletop would you use to help you find the height of the table from the ground? Explain., then find the height of the table.

Multiple Choice

1. A ball initially moves 5 m/s right (positive direction). After 3 seconds, the ball starts accelerating 1 m/s/s to the left indefinitely. Which description best fits the graphs of Velocity v. Time and Acceleration v. Time? Velocity v. Time column listed first, then Acceleration v. Time is second

0 slope at 5 m/s until t=3, constant slope of -1 after t=3

0 slope at 5 m/s/s until t=3, then slope of 0 at 1 m/s/s after t=3

0 slope at 5 m/s until t=3, constant slope of -1 after t=3

0 slope at X-axis until t=3, then slope of 0 at -1 m/s/s after t=3

0 slope at 5 m/s until t=3, 0 slope at X-axis after t=3

0 slope at X-axis throughout

Slope of 5 from t=0 to t=3, constant slope of 1 after t=3

0 slope at X-axis until t=3, then slope of 0 at -1 m/s/s after t=3

Slope of 5 from t=0 to t=3, constant slope of -1 after t=3

0 slope at x-axis throughout

A: Row 1
B: Row 2
C: Row 3
D: Row 4
E: Row 5

2. A boy runs off a 10 ft. high cliff into a pool with an inner tube floating 5m away. With what velocity will he need to run with so he lands in the inner tube? (I hope you meant 10m - Kevin)
A. 1 m/s
B. 2 m/s
C. 3 m/s
D. 4 m/s
E. 5 m/s

3. A different boy from the one in problem 2 walked 4m 45 degrees east of north, then 4m 45 degrees South from east. But then, he realized he dropped his wallet, so he turned around 180 degrees and walked 4 meters again to get back to where he just was. He went to pick up his wallet, but it was gone. He vowed to never take such precisely measured steps again. Anyway, what was the man's total displacement and total distance travelled? Total DistanceTotal Displacement
A. 4m, 4m 45 degrees E of N
B. 12m, 12m 45 degrees E of N
C. 12m, 4m 45 degrees E of N
D. 12m, 45 degrees E of N 4m
E. 4m, 45 degrees E of N 12m

4. A paper airplane is thrown off the roof of a 80 foot tall building. Its velocity as a function of time can be written as v= 3t^2+5t. What is the airplane's acceleration at 7 seconds?
A. 182 m/s/s
B. 156 m/s/s
C. 91 m/s/s
D. 47 m/s/s
E. 18 m/s/s

5. A motorboat travels at 8 m/s to the west against the 3 m/s current of the river flowing east. A person walks parallel to the shore at 5 m/s to the west. What is the velocity of the motorboat and the person on the shore relative to the river's current?

Kinematics in One and Two Dimensions- Displacement, Velocity, Acceleration
- Displacement: the distance that an object travels with respect to its origin.
- Velocity: vector measurement of the how far an object is displaced over a period of time
- Acceleration: change in velocity (can be in speed

- Change in distance in the presence/absence of acceleration
- Without acceleration
- Constant velocity --> Constant change in distance

- With Acceleration
- Distance changes exponentially while velocity changes at a constant rate
- (Think of the graphs: slope of D v. T graph is velocity; slope of V v. T graph is acceleration. Acceleration makes exponential D v. T graph)

- +/- values for each of these
- Subjective references of direction
- Displacement:
- +: Object moved closer to/away from the origin
- -: Object moved in a direction that is opposite to what is defined as the positive direction

- Velocity
- +: Object moving in positively defined direction
- -: Object moving in negatively defined direction

- Acceleration
- +: Velocity changes to become more positive
- -: Velocity changes to become more negative
- Acceleration and Velocity in same direction --> speed increases
- Acceleration and Velocity in different direction --> speed decreases

- Graphs versus Time
- Displacement
- 0 slope: Object stays at rest for period of time
- + slope: Object moves from origin in positively defined direction
- - slope: Object moves from origin in negatively defined direction
- Denotes constant velocity of slope is constant
- Denotes acceleration if graph is exponential function

- Velocity
- 0 slope: Object moves with constant velocity
- +/- slope: Object accelerates
- Denotes constant acceleration if slope of a line is constant
- Object increases speed if the line moves away from the X-axis
- Object decreases speed if the line moves toward the X-axis

- Acceleration
- 0 slope: Object moves with constant acceleration
- +/- slope: Object moves with changing acceleration
- Acceleration and Velocity in same direction --> speed increases
- Acceleration and Velocity in different direction --> speed decreases

Relative MotionORdirection!)- Put simply, it relies on the use of reference points
- If the motion of anything confuses you, just reimagine the situation on a train (usually works out)

Scale Drawings- Be sure to be consistent with scale
- 35 degrees E of N notation (move 35 degrees east from North)

Vectors/trigMajor Equations/Constants:

V= d/t (also d =vt or t=d/v)

d=Vit+1/2at^2

Vf=Vi+at

(Vf)^2=(Vi)^2+2ad

Investigation:An investigation that we did for this unit was the "Shoot Your Grade" experiment. This investigation caused our class to apply our knowledge of kinematics and projectile motion to predict the displacement of a marble shot from a catapult. First, each group was given a specific angle and setting for power on the catapult. The groups then recorded the time it took for the marble to be fired from the catapult, and using the diameter of the marble as "d" (I forget what it was), the initial velocity of the marble was found by dividing by the newly found time, "t". Using components of a right triangle, one could use this velocity to be the hypotenuse of a right triangle, shot from x degrees from the horizontal. However, since the angle was initially given to each group, one could use a trig function to find the velocity of the X and Y components (v cos x for horizontal, v sin x for vertical).

After all the basic information is found, there are a number of ways to approach this problem. The way that I did this was by first calculating the total amount of time that the marble was airborne and by multiplying this time by the horizontal velocity as calculated from the components. This would work assuming that there are ideal conditions (No air resistance, meaning that there would be no force in the X-direction acting on the marble. The marble, therefore, would travel with a constant velocity in the X direction.). To find the time, one would need to start by finding the time it took the marble to reach its maximum height. Then, by using the maximum height, the remaining time would be would by calculating how much time it would take for the marble to fall to the ground from it's maximum height. The velocity in the X component would already be calculated.

The first step one would take would be to find the maximum height of the marble. Since the initial velocity in the Y (vertical) direction was already found through the use of components, and since we know the final velocity in this direction is 0 m/s/s and acceleration the ball would face due to gravity (9.8 m/s/s), the time it would take the marble to reach this height can be calculated by using the formula Vf=Vi+at, which would be rearranged to be t = (Vf-Vi)/a. An important thing to note is that the acceleration due to gravity should be denoted with a negative sign to show that it is

against the current motionof the projectile. This opposition should be shown either in this way (which is probably easier) or by changing to initial velocity to a negative number. Once the time is found, there is enough information to find the height by using the formula d=Vit+1/2at^2, where d is the height (vertical distance, technically). Again, there should be representation of the acceleration due to gravity opposing the initial motion of the marble by using a negative sign. Once this value is found, the last thing to do to measure the height is to measure the distance of the catapult from the floor and add this value to the calculated value (since the calculations show the distance from the catapult, not the floor).After finding this height, one would need to calculate the time it takes for the marble to fall to the ground. The initial velocity would be zero since, in the Y component, there is no velocity, (though the marble is still moving horizontally). The height and acceleration due to gravity are already known, so the equation (Vf)^2=(Vi)^2+2ad can be used to find the velocity of the marble right before it hits the ground (to get Vf by itself, though, you would need to take the square root of both sides of the equation first). With this velocity, the equation Vf=Vi+at can be rearranged to find the time it takes for the marble to acquire this speed [t = (Vf-Vi)/a]. The next thing to do is to add this time with the previously found time it takes for the marble to reach its maximum height to get the total amount of time that the marble was airborne.

With the total time the marble is in the air and the unchanging velocity of the marble in the X component, one can use the equation d=vt to calculate the horizontal distance at which the marble would hit the ground. As stated before, this equation assumes ideal conditions. To account for air resistance, one could make a rough estimate by subtracting between 5% and 10% of the calculated distance.

Open ResponseQuestionA ball rests on top of an inclined plane. It is given a slight push and rolls down the plane and onto a frictionless table top. The ball then rolls off the table and falls onto the ground.

A. At each stage of the marble's motion, state whether there is a constant acceleration, a constant velocity, or neither. Explain your reasoning for each.

B. The ball accelerates a.75 m/s/s down the ramp, beginning at the table with a speed of 6 m/s. How long is the ramp?

C. How long did it take the ball to attain this velocity?

D. What would the velocity be if the ball were initially pushed to go 2 m/s down the ramp? How long would it take the ball to reach 6 m/s?

E. The ball rolls off the table and is airborne for 2.5 seconds before it hits the ground. Which calculated velocity of the tabletop would you use to help you find the height of the table from the ground? Explain., then find the height of the table.

Multiple Choice1. A ball initially moves 5 m/s right (positive direction). After 3 seconds, the ball starts accelerating 1 m/s/s to the left indefinitely. Which description best fits the graphs of Velocity v. Time and Acceleration v. Time?

Velocity v. Timecolumn listed first, thenAcceleration v. Timeis secondB: Row 2

C: Row 3

D: Row 4

E: Row 5

2. A boy runs off a 10

ft. high cliff into a pool with an inner tube floating 5m away. With what velocity will he need to run with so he lands in the inner tube? (I hope you meant 10m - Kevin)A. 1 m/s

B. 2 m/s

C. 3 m/s

D. 4 m/s

E. 5 m/s

3. A different boy from the one in problem 2 walked 4m 45 degrees east of north, then 4m 45 degrees South from east. But then, he realized he dropped his wallet, so he turned around 180 degrees and walked 4 meters again to get back to where he just was. He went to pick up his wallet, but it was gone. He vowed to never take such precisely measured steps again. Anyway, what was the man's total displacement and total distance travelled?

Total DistanceTotal DisplacementA. 4m, 4m 45 degrees E of N

B. 12m, 12m 45 degrees E of N

C. 12m, 4m 45 degrees E of N

D. 12m, 45 degrees E of N 4m

E. 4m, 45 degrees E of N 12m

4. A paper airplane is thrown off the roof of a 80 foot tall building. Its velocity as a function of time can be written as v= 3t^2+5t. What is the airplane's acceleration at 7 seconds?

A. 182 m/s/s

B. 156 m/s/s

C. 91 m/s/s

D. 47 m/s/s

E. 18 m/s/s

5. A motorboat travels at 8 m/s to the west against the 3 m/s current of the river flowing east. A person walks parallel to the shore at 5 m/s to the west. What is the velocity of the motorboat and the person on the shore relative to the river's current?

A: Row 1

B: Row 2

C: Row 3

D: Row 4

E: Row 5

Motorboat Velocity (m/s)Person's Velocity (m/s)