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Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins). Correct the wrong responses you had on the exam by attaching a sheet of paper to it with the corrections/ explanations.

Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

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Page 1: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Aim 34: Review!!!Do Now: Due to unforeseen

circumstances Exam I will not be graded as an exam!

Please take your exam back and a project contract (1 mins).

Correct the wrong responses you had on the exam by attaching a sheet of

paper to it with the corrections/ explanations.

Page 2: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Displacement – has magnitude and direction

Which is a vector quantity?speed masswork displacement

• V 1 – Q 1• V2 – Q 7• V 3 – Q 1

Page 3: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A race car starting from rest accelerates uniformly at a rate of 4.90 meters per second2. What is the car’s speed after it has traveled 200. meters?

• G: vi = 0 m/s , a = 4.90 m/s2, d = 200 m

• U: vf = ?

• E: vf2 = vi

2 + 2 ad

• S: vf2 = 0m/s + 2 (4.9 m/s2 ) (200m)

• vf2 = 1960 m2/s2

• vf= 44.27188724 m/s

• S: vf= 44.3 m/s

• V 1 – Q 2• V2 – Q 13• V 3 – Q 2

Page 4: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A ball is thrown straight downward with a speed of 0.50 meter per second from a height of 4.0 meters. What is the speed of the ball 0.70 second after it is released? [Neglect friction.]

• G: vi = 0.50 m/s , t = 0.70 s, d = 4.0m, a = 9.8 m/s2,

• U: vf = ?

• E: vf = vi + at

• S: vf = 0.50m/s + (9.8 m/s2 ) (0.70 s)

• vf = 7.36 m/s

• S: vf= 7.4 m/s

• V 1 – Q 3• V2 – Q 3• V 3 – Q 3

Page 5: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• In the diagram a 20.-newton force due north and a 20.-newton force due east act concurrently on an object, as shown in the diagram.

• The additional force necessary to bring the object into a state of equilibrium is

• a2 + b2 = c 2

• 20 2 + 20 2 = c2

• c = 28 N (SW)

Page 6: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

5. Which object has the greatest inertia?a 5.00-kg mass moving at 10.0 m/sa 10.0-kg mass moving at 1.00 m/sa 15.0-kg mass moving at 10.0 m/sa 20.0-kg mass moving at 1.00 m/s

• The one with the biggest mass is the 20 Kg mass. Disregard all other information than the mass for inertia problems.

• Inertia Mass

• V 1 – Q 5• V2 – Q 15• V 3 – Q 5

Page 7: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• 6. As a car is driven south in a straight line with decreasing speed, the acceleration of the car must bedirected northwarddirected southward zero constant, but not zero

• velocity and acceleration must be in opposite directions if you are slowing down.

Page 8: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A baseball dropped from the roof of a tall building takes 3.1 seconds to hit the ground. How tall is the building? [Neglect friction.]

• G: vi = 0 m/s , a = 9.8 m/s2, t = 3.1 s• U: d = ? • E: d = vi t + ½ a t2

• S: d = (0m/s)(3.1s) + ½ (9.8 m/s2 ) (3.1s)2

• d = 47.089 m• S: d = 47 m • V 1 – Q 7

• V2 – Q 1• V 3 – Q 7

Page 9: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

8. Which object has the greatest inertia?a falling leafa softball in flighta seated high school studenta rising helium-filled toy balloon

• The one with the biggest mass is the high school student.

Page 10: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• On a highway, a car is driven 80. Kilometers during the first 1.00 hour of travel, 50. kilometers during the next 0.50 hour, and 40. kilometers in the final 0.50 hour. What is the car’s average speed for the entire trip?

• G: d1 = 80 km, d2 = 50 km, d3 = 40 km, t1 = 1hr, t2 = 0.50 hr, t3 = 0.50 hr

• U: average v = ??• E: average v = d/t • S: average v = (80km + 50km+ 40 km)/(2hrs) • V = 170 km/ 2hr• S: v= 85 km/ hr

• V 1 – Q 9• V2 – Q 16• V 3 – Q 9

Page 11: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A high-speed train in Japan travels a distance of 300. kilometers in 3.60 × 103 seconds. What is the average speed of this train?

• G: t = d = 300 km, 3.60 × 103 s• U: average v = ? • E: average v = d/t • S: v = (300km )(1000m/km)/ 3.60 × 103 s• v = 83.3333333 m/s• S: v= 83.3 m/s

Page 12: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• G: vi = 70. m/s , vf = 0m/s, a = 2.0 m/s2

• U: d = ? • E: vf

2 = vi2 + 2 ad

• S: 0m/s = (70m/s)2 + 2 (2.0 m/s2 ) d• 4900 m2/s2 = 4 m/s2 d• 1225m= d• S: 1200 m = d

• V 1 – Q 11• V2 – Q 24• V 3 – Q 11

A 747 jet, traveling at a velocity of 70. meters per second north, touches down on a runway. The jet slows to rest at the rate of 2.0 meters per second2.

• Calculate the total distance the jet travels on the runway as it is brought to rest. [Show all work, including the equation and substitution with units.] [2]

Page 13: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A 747 jet, traveling at a velocity of 70. meters per second north, touches down on a runway. The jet slows to rest at the rate of 2.0 meters per second2.

• On the diagram, point P represents the position of the jet on the runway. Beginning at point P, draw a vector to represent the magnitude and direction of the acceleration of the jet as it comes to rest. Use a scale of 1.0 centimeter = 0.50 meter/second2. [2]

4 cm

2 cm for every m/s 2

You have 2 m/s 2 2x2 = 4

Page 14: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• Which graph best represents the motion of an object in equilibrium?

• Think about what equilibrium means – no acceleration – constant velocity.

• Same slope for a x-t graph or a horizontal in v-t graphs.

Page 15: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A car traveling west in a straight line on a highway decreases its speed from 30.0 meters per second to 23.0 meters per second in 2.00 seconds. The car’s average acceleration during this time interval is

• G: vi = 30.0 m/s , vf = 23.0 m/s , t = 2.00 s• U: a = ? • E: a = (vf - vi )/t• S: a = (23.0m/s- 30m/s)/(2.00s)• a = -7m/s / 2 s• S: d = -3.5 m/s2

• 3.5 m/s2 in the opposite direction or eastward.

• V 1 – Q 14• V2 – Q 2• V 3 – Q 14

Page 16: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• In a race, a runner traveled 12 meters in 4.0 seconds as she accelerated uniformly from rest. The magnitude of the acceleration of the runner was

• G: vi = 0.0 m/s , d = 12 m , t = 4.0 s• U: a = ? • E: d = vi t + ½ a t2

• S: 12 m = (0m/s)(4.0s) + ½ a (4.0s)2

• 12m = a ( 16s2) /2• 12m = a(8s2)• S: 1.5m/s2 = a

• V 1 – Q 15• V2 – Q 14• V 3 – Q 15

Page 17: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Which situation represents a person in equilibrium?a child gaining speed while sliding down a slidea woman accelerating upward in an elevatora man standing still on a bathroom scalea teenager driving around a corner in his car

• Think about what equilibrium means – no acceleration – constant velocity. • V 1 – Q 16

• V2 – Q 5• V 3 – Q 16

Page 18: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• The diagram below shows a compressed spring between two carts initially at rest on a horizontal, frictionless surface. Cart A has a mass of 2 kilograms and cart B has a mass of 1 kilogram. A string holds the carts together.

• The string is cut and the carts move apart. Compared to the magnitude of the force the spring exerts on cart A, the magnitude of the force the spring exerts on cart B is the same Newton 3rd law pairs!

Page 19: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• In a drill during basketball practice, a player runs the length of the 30.-meter court and back. The player does this three times in 60. seconds.

The magnitude of the player’s total displacement after running the drill is

0.0 m

He goes back to where he started at!

Page 20: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• In a drill during basketball practice, a player runs the length of the 30.-meter court and back. The player does this three times in 60. seconds.

The average speed of the player during the drill is

3.0 m/s

(30 m x 6) / 60 s = 180m/ 60s = 3 m/s

Page 21: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A number of 1.0-newton horizontal forces are exerted on a block on a frictionless, horizontal surface. Which top-view diagram shows the forces producing the greatest magnitude of acceleration of the block?

Page 22: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A car, initially traveling east with a speed of 5.0 meters per second, is accelerated uniformly at 2.0 meters per second2 east for 10. Seconds along a straight line. During this 10.-second interval the car travels a total distance of

• G: vi = 5.0 m/s , a = 2.0 m/s2 , t = 10. s• U: d = ? • E: d = vi t + ½ a t2

• S: d = (5.0m/s)(10.s) + ½ (2.0 m/s2)(4.0s)2

• d = 50m + 16m• d = 66m• S: d = 60 m

• V 1 – Q 21• V2 – Q 16• V 3 – Q 21

Page 23: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Which object has the greatest inertia?a 0.010-kg bullet traveling at 90. m/sa 30.-kg child traveling at 10. m/s on her bikea 490-kg elephant walking with a speed of 1.0 m/sa 1500-kg car at rest in a parking lot

• The one with the biggest mass is the parked car.

Page 24: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Four forces act concurrently on a block on a horizontal surface as shown in the diagram.

As a result of these forces, the block

• accelerates to the left

Page 25: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

• A 100.0-kilogram boy and a 50.0-kilogram girl, each holding a spring scale, pull against each other as shown in the diagram below.

The graph below shows the relationship between the magnitude of the force that the boy applies on his spring scale and time.

Page 26: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Which graph best represents the relationship between the magnitude of the force that the girl applies on her spring scale and time?

Newton’s 3rd law Pair!

Page 27: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Bonus Rounds I

• Vector: A physical quantity with magnitude (size) and direction.

• Velocity: The change in displacement in an interval of time.

• Force: Any action that changes the motion of an object.

Page 28: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Bonus Rounds II

• Newton’s 3rd law: A force on B by A = the force B applies to A in the opposite direction.

• Law of Inertia: Objects resist a change in their velocity based on their mass.

Page 29: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Bonus Rounds III

• Draw the free-body diagram for a skier who is moving at a constant velocity to the right.

Normal force

Force of gravity

Page 30: Aim 34: Review!!! Do Now: Due to unforeseen circumstances Exam I will not be graded as an exam! Please take your exam back and a project contract (1 mins)

Bonus Round IV• If Leito kicked a soccer ball on a level field has an

initial vertical velocity component of 15.0 meters per second. Assuming the ball lands at the same height from which it was kicked, what is the total time the ball is in the air? [Neglect friction.]

• G: vi = 15.0 m/s , vf = 0 m/s, a = 9.8 m/s2, • U: t = ? (round trip) x 2• E: vf = vi + at • S: 0m/s = 15 m/s + 9.8 m/s2 t • S: -15m/s = (9.8 m/s2 ) t• 1.5306 s = t x 2 = 3.06 s• S: t = 3.1 s