Presentation on theme: "Unit 3 Kinematics Equations"— Presentation transcript:
1Unit 3 Kinematics Equations Objectives:Learn the 4 motion equations for 1 dimensional motionwhen acceleration is constant
2Motion at Constant Acceleration 1. Kinematics Equation 1Motion at Constant Accelerationa =ΔvΔtbut since "Δ" means changeΔv = v - vo andΔt = t - toif we always let to = 0, Δt = tSolving for "v"This equation tells us how an object's velocity changes as a function of time.a =v - vo∆ta∙∆t = v - voV - vo = a∙∆t𝐕= 𝑽 𝟎 +𝒂∙∆t
3Problem-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.
4Eg1.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∆tStep3: substitute the related variables into equationV=0+ (4.0m/s2)(6.0s)=24m/s
5eg2Starting 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∆tStep3: substitute the related variables into equationV=0+ (8.0m/s2)(9.0s)=72m/s
6Step1: 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 motionVaStep2: 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∆tStep4: substitute the related variables into equationV=5.0m/s+ (-1.5m/s2)(4.0s)=-1.0m/s
7Step1: 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 motionVaStep2: 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∆tStep4: substitute the related variables into equationV=−3.0m/s+ (2.5m/s2)(9.0s)=+19.5m/s
8Eg5.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?
9Eg6.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?
10Eg7.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?
11Eg8.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?
13Motion 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 ) 2𝑉 = (∆𝑥) ∆𝑡And we learned earlier that(𝑉+ 𝑉 0 ) 2 = (∆𝑥) ∆𝑡∆𝑥= (𝑉+ 𝑉 0 )∆𝑡 2Some 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
1452Starting from rest you accelerate to 20 m/s in 4.0s. What is your average velocity?
1553Starting with a velocity of 12 m/s you accelerate to 48 m/s in 6.0s. What is your average velocity?
16Step1: interpret the motion 54Starting 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 motionVaStep2: 𝑉 0 =12.0m/s, V=48𝑚/𝑠, ∆𝑡=6.0𝑠, ∆𝑥=?step3: this is motion problem, use equationStep4: substitute the related variables into equation∆𝑥= (12𝑚/𝑠+48𝑚/𝑠) 6𝑠 × 6.0𝑠 =60𝑚∆𝑥= (𝑉+ 𝑉 0 ) 2 ∆𝑡
18Motion at Constant Acceleration 𝑉 = (∆𝑥) ∆𝑡v =v + vo2v = vo + a∆tΔxtv ∆=x - xo = ½ (v + vo)∆tx - xo = ½v∆t + ½vo∆𝒕x = xo + ½vo∆𝒕 + ½v∆tx = xo + ½vo∆𝒕 + ½(vo + a∆t)∆tx = xo + ½vo ∆𝒕 + ½vo∆𝒕 + ½a(∆𝒕 )2x = xo + vo ∆𝒕 + ½a(∆𝒕 )2We 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
19Motion at Constant Acceleration Graphical ApproachIf 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 = v0tΔxtv(m/s)t (s)A = lw
20Motion at Constant Acceleration Graphical Approachv(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) = ½at2ΔvtA = ½bh
21Motion at Constant Acceleration Graphical ApproachTherefore, the area under a velocity vs. time graph is displacement. It can be calculated by combining the previous two results.A = Δx = v0t + ½at2x - x0 = v0t + ½at2x = x0 + v0t + ½at2v(m/s)t (s)½at2v0t
22Step1: interpret the motion eg9An 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 motionVaStep2: 𝑉 0 =0m/s, a= 3.0𝑚 𝑠2 , ∆𝑡=30.0𝑠, 𝑉=𝑢𝑛𝑘𝑛𝑜𝑤𝑛,∆𝑥=?step3: this is motion problem, use equationStep4: 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
23eg10A 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?
24Step1: interpret the motion eg11A 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 motionVaStep2: 𝑉 0 =0m/s, a= 20𝑚 𝑠2 ,𝑉=𝑛𝑜, ∆𝑥=300𝑚, 𝑡=?step3: this is motion problem, use equationStep4: substitute the related variables into equation300=0× 𝑡 𝑡 2solve t from above equation: 𝑡= ± 30 =±5.5 𝑠𝑒𝑐Step5: apply to real life physics, we can only take t=5.5sec.∆𝑥= 𝑉 0 𝑡+ 1 2 𝑎 𝑡 2
25eg12A 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?
26eg13A 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?
27eg14A 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?
29Motion at Constant Acceleration We can also combine these equations so as to eliminate t:Use the relation, 𝑉 𝑎𝑣𝑒 = 𝑉+ 𝑉 when a=constantWe can prove:∆𝒕= 𝑉− 𝑉 0 𝑎𝑉− 𝑉 0 =𝑎∆𝒕𝑥− 𝑥 0 = 𝑉+ 𝑉 0 2 ∆𝒕= 𝑉+ 𝑉 𝑉− 𝑉 0 𝑎𝑽 𝟐 = 𝑽 𝟎 𝟐 +𝟐𝒂(𝒙− 𝒙 𝟎 )𝑽 𝟐 = 𝑽 𝟎 𝟐 +𝟐𝒂(∆𝒙)
30Step1: interpret the motion eg15A car accelerates from rest to 30m/s while traveling a distance of 20m; what was its acceleration?Step1: interpret the motionVaStep2: 𝑉 0 =0m/s, V= 30𝑚 𝑠 ,∆𝑥=20𝑚, ∆𝑡=𝑛𝑜, 𝑎=?step3: this is motion problem, use equationStep4: substitute the related variables into equation302=0+2𝑎 20.0solve t from above equation: a= +22.5𝑚/𝑠2𝑉 2 = 𝑉 𝑎∆𝑥
31eg16You accelerate, from rest, at 10m/s2 for a distance of 100m; what is your final velocity?
32Step1: interpret the motion eg17Beginning 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 motionVaStep2: 𝑉 0 =25m/s, a=2.0𝑚/𝑠2,∆𝑥=20𝑚, 𝑉=?step3: this is motion problem, use equationStep4: substitute the related variables into equation𝑉2=252+2(2.0) 200solve 𝑉=± = ±38 𝑚/𝑠We know the velocity should be positive to the right, v=+38m/s𝑉 2 = 𝑉 𝑎∆𝑥
33eg18You accelerate from 20m/s to 60m/s while traveling a distance of 200m; what was your acceleration?
34eg19A dropped ball falls 8.0m; what is its final velocity?
35eg20A 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?
37Motion 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) 2𝑉 2 = 𝑉 𝑎∆𝑥
38Sample Problem Chapter 2 Final Velocity After Any Displacement A person pushing a stroller starts from rest, uniformlyaccelerating at a rate of m/s2. What is thevelocity of the stroller after it has traveled 4.75 m?
39Sample Problem, continued 1. DefineGiven:v0 = 0 m/s∆𝑡= ?a = m/s2x = 4.75 m𝑉= ?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.
40Sample Problem, continued Chapter 22. PlanChoose 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 .
41Sample Problem, continued Chapter 23. CalculateSubstitute 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. EvaluateThe stroller’s velocityafter accelerating for 4.75 m is 2.18 m/s to the right.
42Motion at Constant Acceleration The change in velocity during first 12 seconds is equivalent to the shadowed area(4m x 12s = 48m).ss2The change in velocity during first 12 seconds is 48 m/s.
4345The following graph shows acceleration as a function of time of a moving object. What is the change in velocity during first 10 seconds?
44Acceleration Due to Gravity Free Fall:Acceleration Due to Gravity
45Free FallAll 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/s2Keep 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 isa= −𝒈=−9.8 m/s2Fact:The acceleration due to gravity on the surface of the Moon is m/s2, about 16.6% that on Earth's surface.
46What happens at the top?It stops momentarily.v = 0g = -9.8 m/s2What happens when it goes down?It speeds up(negative acceleration)g = -9.8 m/s2What happens when it goes up?It slows down.(negative acceleration)g = -9.8 m/s2What happens when it lands?An object is thrown upward with initial velocity, voIt returns with itsoriginal velocity.
48On the way up:On the way down:v0vav1at = 0 st = 3 sv2at = 2 sv1t = 1 sv2aav1v2t = 1 sv2t = 2 savav0v1t = 3 st = 0 sv
49v0 = 0g = -9.8 m/s2For any object drop from certain height, this is what the velocity vs time graph looks like.v(m/s)t (s)
50v(m/s)t (s)An object is thrown upward with initial velocity, voIt stops momentarily.v = 0g = -9.8 m/s2It returns with itsoriginal 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.
5146A ball is dropped from rest and falls (do not consider air resistance). Which is true about its motion?cAacceleration is constantcBvelocity is constantCvelocity is decreasingcDacceleration is decreasingc
52Jason hits a volleyball so that it moves with an initial Sample ProblemFalling ObjectJason hits a volleyball so that it moves with an initialvelocity of 6.0 m/s straight upward. If the volleyballstarts from 2.0 m above the floor, how long will it bein the air before it strikes the floor?Choose upward as positive direction and 2.0m height as origin of y –axis :0m-2.0m
53Sample Problem, continued 1. DefineGiven: Unknown:v0 = +6.0 m/s t = ?a = –g = –9.81 m/s2y = –2.0 mDiagram:Place the origin at theStarting point of the ball(y0 = 0 at t0 = 0).
54Sample Problem, continued 2. PlanChoose 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∆y𝑉=𝑉 0 +a∆tRearrange 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
55Sample Problem, continued 3. CalculateSubstitute 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.
56Sample Problem, continued Next, use this value of vf in the second equation to solve for ∆t.4. EvaluateThe solution, 1.50 s, is a reasonable amount of time for the ballto be in the air.
5747An 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?
5848A 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?
5949A 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?
6050An 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?
6151A rocket is fired straight up from the ground. It returns to the ground 10 seconds later. What was it's launch speed?