Download presentation

Presentation is loading. Please wait.

1
**Unit 3 Kinematics Equations**

Objectives: Learn the 4 motion equations for 1 dimensional motion when acceleration is constant

2
**Motion at Constant Acceleration**

1. Kinematics Equation 1 Motion at Constant Acceleration a = Δv Δt but since "Δ" means change Δv = v - vo and Δt = t - to if we always let to = 0, Δt = t Solving for "v" This equation tells us how an object's velocity changes as a function of time. a = v - vo ∆t a∙∆t = v - vo V - vo = a∙∆t 𝐕= 𝑽 𝟎 +𝒂∙∆t

3
**Problem-Solving Strategy**

The strategy involves the following steps: Construct an informative diagram of the physical situation. Identify and list the given information in variable form. Identify and list the unknown information in variable form. Identify and list the equation that will be used to determine unknown information from known information. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. Check your answer to insure that it is reasonable and mathematically correct.

4
Eg1. Starting from rest, you accelerate at 4.0 m/s2 for 6.0s. What is your final velocity? Step1: 𝑉 0 =0, 𝑎=4.0𝑚/𝑠2, ∆𝑡=6.0𝑠, 𝑉=? step2: this is motion problem, use equation V= 𝑉 0 +a∆t Step3: substitute the related variables into equation V=0+ (4.0m/s2)(6.0s)=24m/s

5
eg2 Starting from rest, you accelerate at 8.0 m/s2 for 9.0s. What is your final velocity? Step1: 𝑉 0 =0, 𝑎= 8.0𝑚 𝑠2 , ∆𝑡=9.0𝑠, ∆𝑥=𝑢𝑛𝑘𝑛𝑜𝑤𝑛, 𝑉=? step2: this is motion problem, use equation V= 𝑉 0 +a∆t Step3: substitute the related variables into equation V=0+ (8.0m/s2)(9.0s)=72m/s

6
**Step1: interpret the motion**

Eg3. You have an initial velocity of 5.0 m/s. You then experience an acceleration of -1.5 m/s2 for 4.0s; what is your final velocity? Step1: interpret the motion V a Step2: 𝑉 0 =5.0m/s, 𝑎=− 1.5𝑚 𝑠2 , ∆𝑡=4.0𝑠, ∆𝑥=𝑢𝑛𝑘𝑛𝑜𝑤𝑛, 𝑉=? step3: this is motion problem, use equation V= 𝑉 0 +a∆t Step4: substitute the related variables into equation V=5.0m/s+ (-1.5m/s2)(4.0s)=-1.0m/s

7
**Step1: interpret the motion**

Eg4. You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what is your final velocity? Step1: interpret the motion V a Step2: 𝑉 0 =-3.0m/s, 𝑎=+ 2.5𝑚 𝑠2 , ∆𝑡=9.0𝑠, ∆𝑥=𝑢𝑛𝑘𝑛𝑜𝑤𝑛, 𝑉=? step3: this is motion problem, use equation V= 𝑉 0 +a∆t Step4: substitute the related variables into equation V=−3.0m/s+ (2.5m/s2)(9.0s)=+19.5m/s

8
Eg5. How much time does it take to accelerate from an initial velocity of 20m/s to a final velocity of 100m/s if your acceleration is 1.5 m/s2?

9
Eg6. How much time does it take to come to rest if your initial velocity is 5.0 m/s and your acceleration is -2.0 m/s2?

10
Eg7. An object accelerates at a rate of 3 m/s2 for 6 s until it reaches a velocity of 20 m/s. What was its initial velocity?

11
Eg8. An object accelerates at a rate of 1.5 m/s2 for 4 s until it reaches a velocity of 10 m/s. What was its initial velocity?

12
2. Kinematics Equation 2

13
**Motion at Constant Acceleration**

If velocity is changing at a constant rate, the average velocity is just the average of the initial and final velocities. 𝑉 = (𝑉+ 𝑉 0 ) 2 𝑉 = (∆𝑥) ∆𝑡 And we learned earlier that (𝑉+ 𝑉 0 ) 2 = (∆𝑥) ∆𝑡 ∆𝑥= (𝑉+ 𝑉 0 )∆𝑡 2 Some problems can be solved most easily by using these two equations together. It shows displacement as function of time. 𝑥− 𝑥 0 = (𝑉+ 𝑉 0 ) 2 ∆𝑡 ∆𝑥= (𝑉+ 𝑉 0 ) 2 ∆𝑡

14
52 Starting from rest you accelerate to 20 m/s in 4.0s. What is your average velocity?

15
53 Starting with a velocity of 12 m/s you accelerate to 48 m/s in 6.0s. What is your average velocity?

16
**Step1: interpret the motion**

54 Starting with a velocity of 12 m/s you accelerate to 48 m/s in 6.0s. Using your previous answer, how far did you travel in that 6.0s? Step1: interpret the motion V a Step2: 𝑉 0 =12.0m/s, V=48𝑚/𝑠, ∆𝑡=6.0𝑠, ∆𝑥=? step3: this is motion problem, use equation Step4: substitute the related variables into equation ∆𝑥= (12𝑚/𝑠+48𝑚/𝑠) 6𝑠 × 6.0𝑠 =60𝑚 ∆𝑥= (𝑉+ 𝑉 0 ) 2 ∆𝑡

17
3. Kinematics Equation 3

18
**Motion at Constant Acceleration**

𝑉 = (∆𝑥) ∆𝑡 v = v + vo 2 v = vo + a∆t Δx t v ∆ = x - xo = ½ (v + vo)∆t x - xo = ½v∆t + ½vo∆𝒕 x = xo + ½vo∆𝒕 + ½v∆t x = xo + ½vo∆𝒕 + ½(vo + a∆t)∆t x = xo + ½vo ∆𝒕 + ½vo∆𝒕 + ½a(∆𝒕 )2 x = xo + vo ∆𝒕 + ½a(∆𝒕 )2 We can combine these three equations to derive an equation which will directly tell us the position of an object as a function of time. ∆𝑥= 𝑉 0 ∙∆𝑡+ 1 2 𝑎(∆ 𝑡) 2

19
**Motion at Constant Acceleration**

Graphical Approach If the area under the graph is length x width (A = lw), then: A = v0t Since we know that v = , then area is really Δx. A = Δx = v0t Δx t v (m/s) t (s) A = lw

20
**Motion at Constant Acceleration**

Graphical Approach v (m/s) t (s) If the area under this graph is ½ base x height, then: A = ½ t Δv Since we know that a = , Δv = at. A = Δx = ½t(at) = ½at2 Δv t A = ½bh

21
**Motion at Constant Acceleration**

Graphical Approach Therefore, the area under a velocity vs. time graph is displacement. It can be calculated by combining the previous two results. A = Δx = v0t + ½at2 x - x0 = v0t + ½at2 x = x0 + v0t + ½at2 v (m/s) t (s) ½at2 v0t

22
**Step1: interpret the motion**

eg9 An airplane starts from rest and accelerates at a constant rate of 3.0 m/s2 for 30.0 s before leaving the ground. How far did it move along the runway? Step1: interpret the motion V a Step2: 𝑉 0 =0m/s, a= 3.0𝑚 𝑠2 , ∆𝑡=30.0𝑠, 𝑉=𝑢𝑛𝑘𝑛𝑜𝑤𝑛, ∆𝑥=? step3: this is motion problem, use equation Step4: substitute the related variables into equation ∆𝑥=0× 30.0𝑠 (3.0𝑚/𝑠2) (30.0𝑠) 2 =1350𝑚 ∆𝑥= 𝑉 0 𝑡+ 1 2 𝑎(∆ 𝑡) 2

23
eg10 A Volkswagen Beetle moves at an initial velocity of 12 m/s. It coasts up a hill with a constant acceleration of –1.6 m/s2. How far has it traveled after 6.0 seconds?

24
**Step1: interpret the motion**

eg11 A motorcycle starts out from a stop sign and accelerates at a constant rate of 20 m/s2. How long will it take the motorcycle to go 300 meters? Step1: interpret the motion V a Step2: 𝑉 0 =0m/s, a= 20𝑚 𝑠2 ,𝑉=𝑛𝑜, ∆𝑥=300𝑚, 𝑡=? step3: this is motion problem, use equation Step4: substitute the related variables into equation 300=0× 𝑡 𝑡 2 solve t from above equation: 𝑡= ± 30 =±5.5 𝑠𝑒𝑐 Step5: apply to real life physics, we can only take t=5.5sec. ∆𝑥= 𝑉 0 𝑡+ 1 2 𝑎 𝑡 2

25
eg12 A train pulling out of Grand Central Station accelerates from rest at a constant rate. It covers 800 meters in 20 seconds. What is its rate of acceleration?

26
eg13 A car has a initial velocity of 45 m/s. It accelerates for 4.8 seconds. In this time, the car covers 264 meters. What is its rate of acceleration?

27
eg14 A Greyhound bus traveling at a constant velocity starts to accelerate at a constant 2.0 m/s2. If the bus travels 500 meters in 20 seconds, what was its initial velocity?

28
4. Kinematics Equation 4

29
**Motion at Constant Acceleration**

We can also combine these equations so as to eliminate t: Use the relation, 𝑉 𝑎𝑣𝑒 = 𝑉+ 𝑉 when a=constant We can prove: ∆𝒕= 𝑉− 𝑉 0 𝑎 𝑉− 𝑉 0 =𝑎∆𝒕 𝑥− 𝑥 0 = 𝑉+ 𝑉 0 2 ∆𝒕= 𝑉+ 𝑉 𝑉− 𝑉 0 𝑎 𝑽 𝟐 = 𝑽 𝟎 𝟐 +𝟐𝒂(𝒙− 𝒙 𝟎 ) 𝑽 𝟐 = 𝑽 𝟎 𝟐 +𝟐𝒂(∆𝒙)

30
**Step1: interpret the motion**

eg15 A car accelerates from rest to 30m/s while traveling a distance of 20m; what was its acceleration? Step1: interpret the motion V a Step2: 𝑉 0 =0m/s, V= 30𝑚 𝑠 ,∆𝑥=20𝑚, ∆𝑡=𝑛𝑜, 𝑎=? step3: this is motion problem, use equation Step4: substitute the related variables into equation 302=0+2𝑎 20.0 solve t from above equation: a= +22.5𝑚/𝑠2 𝑉 2 = 𝑉 𝑎∆𝑥

31
eg16 You accelerate, from rest, at 10m/s2 for a distance of 100m; what is your final velocity?

32
**Step1: interpret the motion**

eg17 Beginning with a velocity of 25m/s, you accelerate at a rate of 2.0m/s2. During that acceleration you travel 200m; what is your final velocity? Step1: interpret the motion V a Step2: 𝑉 0 =25m/s, a=2.0𝑚/𝑠2,∆𝑥=20𝑚, 𝑉=? step3: this is motion problem, use equation Step4: substitute the related variables into equation 𝑉2=252+2(2.0) 200 solve 𝑉=± = ±38 𝑚/𝑠 We know the velocity should be positive to the right, v=+38m/s 𝑉 2 = 𝑉 𝑎∆𝑥

33
eg18 You accelerate from 20m/s to 60m/s while traveling a distance of 200m; what was your acceleration?

34
eg19 A dropped ball falls 8.0m; what is its final velocity?

35
eg20 A ball with an initial velocity of 25m/s is subject to an acceleration of -9.8 m/s2; how high does it go before coming to a momentary stop?

36
**5. Mixed Kinematics Problems**

37
**Motion at Constant Acceleration**

We now have all the equations we need to solve constant-acceleration problems. 𝑉=𝑉 0 +a∙∆t ∆𝑥= (𝑉+ 𝑉 0 ) 2 ∙∆𝑡 ∆𝑥= 𝑉 0 ∙∆𝑡+ 1 2 𝑎 (∆𝑡) 2 𝑉 2 = 𝑉 𝑎∆𝑥

38
**Sample Problem Chapter 2 Final Velocity After Any Displacement**

A person pushing a stroller starts from rest, uniformly accelerating at a rate of m/s2. What is the velocity of the stroller after it has traveled 4.75 m?

39
**Sample Problem, continued**

1. Define Given: v0 = 0 m/s ∆𝑡= ? a = m/s2 x = 4.75 m 𝑉= ? Diagram: Choose a coordinate system. The most convenient one has an origin at the initial location of the stroller, as shown above. The positive direction is to the right.

40
**Sample Problem, continued**

Chapter 2 2. Plan Choose an equation or situation: Because the initial velocity, acceleration, and displacement are known, the final velocity can be found using the following equation: Rearrange the equation to isolate the unknown: Take the square root of both sides to isolate vf .

41
**Sample Problem, continued**

Chapter 2 3. Calculate Substitute the values into the equation and solve: Tip: Think about the physical situation to determine whether to keep the positive or negative answer from the square root. In this case, the stroller starts from rest and ends with a speed of 2.18 m/s. An object that is speeding up and has a positive acceleration must have a positive velocity. So, the final velocity must be positive. 4. Evaluate The stroller’s velocity after accelerating for 4.75 m is 2.18 m/s to the right.

42
**Motion at Constant Acceleration**

The change in velocity during first 12 seconds is equivalent to the shadowed area (4m x 12s = 48m). s s2 The change in velocity during first 12 seconds is 48 m/s.

43
45 The following graph shows acceleration as a function of time of a moving object. What is the change in velocity during first 10 seconds?

44
**Acceleration Due to Gravity**

Free Fall: Acceleration Due to Gravity

45
Free Fall All unsupported objects fall towards Earth with the same acceleration. We call this acceleration the "acceleration due to gravity" and it is denoted by g. g = 9.8 m/s2 Keep in mind, ALL objects accelerate towards the earth at the same rate. g is a constant! We choose upward as positive, then the free fall object on earth surface acceleration is a= −𝒈=−9.8 m/s2 Fact: The acceleration due to gravity on the surface of the Moon is m/s2, about 16.6% that on Earth's surface.

46
What happens at the top? It stops momentarily. v = 0 g = -9.8 m/s2 What happens when it goes down? It speeds up (negative acceleration) g = -9.8 m/s2 What happens when it goes up? It slows down. (negative acceleration) g = -9.8 m/s2 What happens when it lands? An object is thrown upward with initial velocity, vo It returns with its original velocity.

47
It stops momentarily. v = 0 a = -9.8 m/s2 It speeds up. (negative acceleration) a = -9.8 m/s2 It slows down. (negative acceleration) a = -9.8 m/s2 An object is thrown upward with initial velocity, vo It returns with its original velocity.

48
On the way up: On the way down: v0 v a v1 a t = 0 s t = 3 s v2 a t = 2 s v1 t = 1 s v2 a a v1 v2 t = 1 s v2 t = 2 s a v a v0 v1 t = 3 s t = 0 s v

49
v0 = 0 g = -9.8 m/s2 For any object drop from certain height, this is what the velocity vs time graph looks like. v (m/s) t (s)

50
v (m/s) t (s) An object is thrown upward with initial velocity, vo It stops momentarily. v = 0 g = -9.8 m/s2 It returns with its original velocity but in the opposite direction. For any object thrown straight up into the air, this is what the velocity vs time graph looks like.

51
46 A ball is dropped from rest and falls (do not consider air resistance). Which is true about its motion? c A acceleration is constant c B velocity is constant C velocity is decreasing c D acceleration is decreasing c

52
**Jason hits a volleyball so that it moves with an initial **

Sample Problem Falling Object Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0 m above the floor, how long will it be in the air before it strikes the floor? Choose upward as positive direction and 2.0m height as origin of y –axis : 0m -2.0m

53
**Sample Problem, continued**

1. Define Given: Unknown: v0 = +6.0 m/s t = ? a = –g = –9.81 m/s2 y = –2.0 m Diagram: Place the origin at the Starting point of the ball (y0 = 0 at t0 = 0).

54
**Sample Problem, continued**

2. Plan Choose an equation or situation: Both ∆t and v are unknown. Therefore, first solve for v using the equation that does not require time. Then, the equation for v that does involve time can be used to solve for ∆t. 𝑉 2 = 𝑉 𝑎∆y 𝑉=𝑉 0 +a∆t Rearrange the equation to isolate the unknown: Take the square root of the first equation to isolate v. The second equation must be rearranged to solve for ∆t. 𝑉=± 𝑉 𝑎∆y ∆𝑡= 𝑉− 𝑉 0 𝑎

55
**Sample Problem, continued**

3. Calculate Substitute the values into the equation and solve: First find the velocity of the ball at the moment that it hits the floor. Tip: When you take the square root to find vf , select the negative answer because the ball will be moving toward the floor, in the negative direction.

56
**Sample Problem, continued**

Next, use this value of vf in the second equation to solve for ∆t. 4. Evaluate The solution, 1.50 s, is a reasonable amount of time for the ball to be in the air.

57
47 An acorn falls from an oak tree. You note that it takes 2.5 seconds to hit the ground. How fast was it going when it hit the ground?

58
48 A rock falls off a cliff and hits the ground 5 seconds later. What velocity did it hit the ground with? How high is the cliff?

59
49 A ball is thrown down off a bridge with a velocity of 5 m/s. What is its velocity 2 seconds later? How far did it fall?

60
50 An arrow is fired straight upward into the air and it reaches its highest point 3 seconds later. What was its velocity when it was fired? What is the maximum height it reached?

61
51 A rocket is fired straight up from the ground. It returns to the ground 10 seconds later. What was it's launch speed?

62
**Check your understanding:**

63
**Acceleration of a Bungee Jump**

Now take what you have just learned and contemplate a question about acceleration in the Veritasium video below: Click here for the video

Similar presentations

OK

© Houghton Mifflin Harcourt Publishing Company The student is expected to: Chapter 2 Section 1 Displacement and Velocity TEKS 4A generate and interpret.

© Houghton Mifflin Harcourt Publishing Company The student is expected to: Chapter 2 Section 1 Displacement and Velocity TEKS 4A generate and interpret.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on conjunctions for grade 5 Ppt on rabindranath tagore in english Cell surface display ppt online Ppt on computer languages of the future Download ppt on reduce reuse recycle Ppt on mammals and egg laying animals birds Ppt on world population day 2011 Ppt on properties of different quadrilaterals Ppt on interview etiquettes of life Ppt on dry cell and wet cellulose