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Unit 3 Kinematics Equations

Unit 3 Kinematics Equations. Objectives: Learn the 4 motion equations for 1 dimensional motion when acceleration is constant. 1. Kinematics Equation 1 Motion at Constant Acceleration. Δ v. a =. Δ t. but since "Δ" means change Δ v = v - v o and Δ t = t - t o

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Unit 3 Kinematics Equations

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  1. Unit 3 Kinematics Equations Objectives: Learn the 4 motion equations for 1 dimensional motion when acceleration is constant

  2. 1. Kinematics Equation 1 Motion at Constant Acceleration Δv a = 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. v - vo a = t at = v - vo V- vo = at 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: =0, step2: this is motion problem, use equation V= 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: =0, step2: this is motion problem, use equation V= Step3: substitute the related variables into equation V=0+ (8.0m/s2)(9.0s)=72m/s

  6. 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: =5.0m/s, step3: this is motion problem, use equation V= Step4: substitute the related variables into equation V=5.0m/s+ (-1.5m/s2)(4.0s)=-1.0m/s

  7. 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: =-3.0m/s, step3: this is motion problem, use equation V= Step4: substitute the related variables into equation V=3.0m/s+ (2.5m/s2)(9.0s)=+19.5m/s

  8. 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? Eg5.

  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. 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? Eg8.

  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. And we learned earlier that Some problems can be solved most easily by using these two equations together. It shows displacement as function of time.

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

  15. 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: =12.0m/s, V step3: this is motion problem, use equation Step4: substitute the related variables into equation

  16. 3. Kinematics Equation 3

  17. Motion at Constant Acceleration v + vo v = 2 Δx v t = v = vo + at x - xo = ½ (v + vo)t x - xo = ½vt + ½vo x = xo + ½vo + ½vt x = xo + ½vo + ½(vo + at)t x = xo + ½vo + ½vo + ½a()2 x = xo + vo + ½a()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.

  18. 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? eg9 Step1: interpret the motion V a Step2: =0m/s, a step3: this is motion problem, use equation Step4: substitute the related variables into equation

  19. 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?

  20. 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: =0m/s, a step3: this is motion problem, use equation Step4: substitute the related variables into equation solve t from above equation: Step5: apply to real life physics, we can only take t=5.5sec.

  21. 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?

  22. 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?

  23. 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?

  24. 4. Kinematics Equation 4

  25. Motion at Constant Acceleration We can also combine these equations so as to eliminate t: Use the relation, when a=constant We can prove:

  26. 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: =0m/s, V step3: this is motion problem, use equation Step4: substitute the related variables into equation solve t from above equation: a

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

  28. 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: =25m/s, a step3: this is motion problem, use equation Step4: substitute the related variables into equation solve = We know the velocity should be positive to the right, v=+38m/s

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

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

  31. 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?

  32. 5. Mixed Kinematics Problems

  33. Motion at Constant Acceleration We now have all the equations we need to solve constant-acceleration problems.

  34. Sample Problem Chapter 2 Final Velocity After Any Displacement A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75 m?

  35. Sample Problem, continued 1. Define Given: v0= 0 m/s v= no a = 0.500 m/s2 x = 4.75 m = ? 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.

  36. 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 .

  37. 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.

  38. Free Fall: Acceleration Due to Gravity

  39. 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= 9.8 m/s2 Fact: The acceleration due to gravity on the surface of the Moon is 1.6249 m/s2, about 16.6% that on Earth's surface.

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

  41. 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 It returns with its original velocity. An object is thrown upward with initial velocity, vo

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

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

  44. 46 A ball is dropped from rest and falls (do not consider air resistance). Which is true about its motion? A acceleration is constant c c B velocity is constant C velocity is decreasing c D acceleration is decreasing c

  45. 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

  46. 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).

  47. Sample Problem, continued 2. Plan Choose an equation or situation: Both ∆t and vare unknown. Therefore, first solve for vusing the equation that does not require time. Then, the equation for vthat does involve time can be used to solve for ∆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.

  48. 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.

  49. 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.

  50. 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? 47

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