Kinematic Notes Part 3 Big 4, Free Fall, & Relative Motion 1 March 17, 2017 Wed Feb 24 Tuesday, February 28th • Score Practice Quizzes • Hand back review sheets • Analyzing Motion • The Big 4 ‐ Equations of Motion • Sample Problems • Practice: p.80 #1, 3; p.89 #4, 6 The Big 4 Equations of Motion a.k.a. The Big 4 v f =v i + at 1 d=v i t + (1/2)at 2 3 d= (v i +v f )t 2 2 v f 2 =v i 2 + 2ad 4 They all mathemagically come from our two simple equations: vave = Δd Δt a= Δv Δt & Deriving #1 v f =v i + at 1 a= Δv Δt a= v f ‐v i Δt at = vf ‐vi at + vi=vf vf =vi + at Start with what we know: what does Δv mean? multiply both sides by t bring vi to the other side rearrange to match above Deriving #2 Δd=vaveΔt Start with what we know: d= (v i +v f )t 2 vave = Δd Δt rearrange to solve for d Δd= (v i +v f )Δt 2 what is vave? 2 Deriving #3 Start with what we know: substitute 1 into 2, replacing vf d=v i t + (1/2)at 2 3 vf =vi + at 1 Δd= (v i +v f )Δt 2 2 Δd= (v i +[v i + at])Δt 2 Δd= (v i +v i + at)Δt 2 Δd= 2v i t + at 2 2 Δd= 2v i t + at 2 2 2 Δd=vit + (1/2)at 2 adding, so drop the inner brackets add vi's and multiply the outside t separate into 2 fractions cancel terms and rewrite as a half Deriving #4 Start with what we know: rearrange 1 to solve for t vf =vi + at 1 Δd= (v i +v f )Δt 2 v f 2 =v i 2 + 2ad 4 t= v f ‐v i a sub this into 2 to get rid of time Δd= (v i +v f )(v f ‐v i ) 2 a 2ad = (vi +vf)(vf ‐vi) multiply both sides by 2a 2ad=vivf +vfvf ‐vivi ‐vivf FOIL the brackets simplify 2ad=vf 2 ‐vi 2 vf 2 =vi 2 + 2ad rearrange to solve for vf 2 2
Transcript
Kinematic Notes Part 3
Big 4, Free Fall, & Relative Motion 1
March 17, 2017
Wed Feb 24
Tuesday, February 28th
• Score Practice Quizzes• Hand back review sheets
• Analyzing Motion• The Big 4 ‐ Equations of Motion• Sample Problems• Practice: p.80 #1, 3; p.89 #4, 6
The Big 4
Equations of Motiona.k.a. The Big 4
vf = vi + at1
d = vit + (1/2)at2 3
d = (vi + vf )t 2
2
vf2 = vi2 + 2ad 4They all mathemagically come from our two simple equations:
sub this into 2 to get rid of time Δd = (vi +vf)(vf ‐ vi) 2 a2ad = (vi +vf)(vf ‐ vi)multiply both sides by 2a
2ad = vivf + vfvf ‐vivi ‐vivf FOIL the brackets
simplify 2ad = vf2 ‐vi2 vf2 = vi2 + 2adrearrange to solve for vf2
2
Kinematic Notes Part 3
Big 4, Free Fall, & Relative Motion 2
March 17, 2017
Sample Problem 1
Sample Problems (p. 77, 78)1. A slight earth tremor causes a large boulder to break free and start rolling down the mountainside with a constant acceleration of 5.2 m/s2. What was the boulder's velocity after 8.5s?
Sample Problem 2
Sample Problems (p. 77, 78)2. A skier is going 8.2 m/s when she falls and starts sliding down the ski run. After 3.0 s, her velocity is 3.1 m/s. Assuming her acceleration was constant, how long after she fell did she finally come to a stop?
Equation Chart
vf = vi + at
d = vit + (1/2)at2
d = (vi + vf )t 2
vf2 = vi2 + 2ad
Which One Should I Use??Equation Missing Variable
d
a
vf
t
Once your Given & Required list is done, check to see if any variables are missing. This will help you choose which equations you may use and which you cannot use right away.
p.80 #1,3; p.89 #4,6
p. 89 #4 & 6
p. 80 #1 & 3
Mar 110:07 AM Thurs, Mar 2
Thursday, March 2nd
• Decide on Test Date
• Analyzing Motion:• Free Fall• Acceleration due to gravity• Sample Problems• Practice Work
Kinematic Notes Part 3
Big 4, Free Fall, & Relative Motion 3
March 17, 2017
Free Fall Free Fall
Watch out if a word problem tells you you're not on Earth!
Free FallGravity exerts a force on objects as they fall to Earth causing them to accelerate at a constant rate (uniform acceleration). The value for the acceleration due to gravity (ag) on Earth:
9.81 m/s2 We usually make [down] negative, so ag is usually negative.
Every planet has its own value for ag based on the planet's mass:
LocationAcceleration due to gravity (m/s2)
EarthMoonMarsJupiter
9.811.643.7225.9
Table 4.4
p. 133
Sample Problems
Sample Problems1. You throw a rock off a cliff, giving it a velocity of 8.3 m/s, straight down. At the instant you let go, your hiking buddy started a stopwatch. You heard the splash when the rock hit the river below, exactly 6.9 s after you threw the rock. How high is the cliff above the river?
Sample Problems
Sample Problems2. Galileo celebrates his discovery of Jupiter's 4 largest moons by joyously throwing his primitive telescope straight up into the air with the initial velocity 3.1 m/s. How long does it take for the telescope to come back and land in his hand?
Free Fall Practice
Practice Problems1. The acceleration due to gravity on the Moon is 1.6 m/s2 (down). If a baseball is thrown with an initial velocity of 4.5 m/s (up), what would its velocity be after 4.0s?
2. A skydiver falling toward the ground accelerates at 3.2m/s2. Calculate his displacement if after 8.0s he attained a velocity of 28 m/s (down). Why isn't his acceleration 9.81 m/s2?
3. A book tips over and falls off a shelf. If the shelf is 3.0m off the ground, how fast will it be moving right before it strikes the floor? How long will it take it hit the floor?
Tues March 14
Tuesday, March 14
• Relative Motion
• Practice Quiz Fri March 17th• Assignment Due Fri March 17th• Unit Test Tues March 21st
Kinematic Notes Part 3
Big 4, Free Fall, & Relative Motion 4
March 17, 2017
Relative Motion
Relative Motion
Relative Motion Ex1
Relative MotionSometimes, the person observing the motion of an object is also moving. This means the reference frame is moving. In these situations, the relative velocity/speed can often be used:
vrelative = vobject ‐ vobserver Ex 1: A car is travelling at 20m/s [East]. A truck is travelling at 30m/s [East]. a) What is the velocity of the truck relative to the car? (What does the car's driver observe?)
b) What is the velocity of the car relative to the truck? (What does the truck's driver observe?)
Relative Motion Ex2
Relative MotionEx 2: Two cars are moving in the same direction at constant velocities. The Car A is traveling 30 km/h and Car B is traveling 40 km/h. If Car B is 4.0 km behind Car A, how long will it take to catch up with Car A?
Relative Motion Ex3
Relative MotionEx 3: Just as the light turns green, a stationary car starts off with an acceleration of 3.00 m/s2. At the same instant, a truck traveling in the same direction with a constant speed of 12.0 m/s passes in the next lane. (a) What distance will the car have to travel in order to catch up with the truck? (b) What speed will the car be going when it catches up with the truck?
