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Unit 2

Unit 2. The Kinematics Equations (1D Equations of Motion). Day #1 Introduction to the Equation of Motion. What are the 5 key equations we have learned thus far?. A) ∆x = x f – x i finding a change in displacement B) ∆v = v f – v i finding a change in velocity

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Unit 2

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  1. Unit 2 The Kinematics Equations (1D Equations of Motion)

  2. Day #1 Introduction to the Equation of Motion

  3. What are the 5 key equations we have learned thus far? • A) ∆x = xf – xi finding a change in displacement • B) ∆v = vf – vi finding a change in velocity • C) s = d/∆t finding speed (distance/time) • D) v = ∆x/∆t finding velocity (displacement/time) • E) a = ∆v/∆t finding acceleration (change in velocity change in time) ¯ ¯

  4. There are 4 new equations for motion. • We’ll start with a = ∆v/∆t • Substitute in ∆v = vf – vi • The formula is now a = vf-vi t • Rearrange to get a∙t = vf - vi ¯ • Remember… • slope of the line = • a = vf-vi t ¯

  5. Example #1 • A car starting from rest accelerates uniformly to a speed of 75 m/s in 12 seconds. What is the car’s acceleration? • First, identify the known values, then plug and chug • a = vf-vi or a∙t = vf - vi t = 75 m/s – 0 12s =+6.25 m/s2 • Vi = 0 • Vf = 75 m/s • t = 12 s • a = ? • *always make a list of your knowns

  6. Example #2 • A spacecraft traveling at 1025 m/s is uniformly accelerated at a rate of 105 m/s2 for 9 seconds. What is the final velocity of the spacecraft? • First, identify the known values, rearrange the formula, then plug and chug • a = vf-vi or a∙t = vf - vi t Vf= Vi + at = 1025 m/s + 9s(105m/s2) = 1970 m/s • Vi = 1025 m/s • Vf = ? • t = 9 s • a = 105 m/s2

  7. Recall: area under the curve on a v/t chart gives us displacement = ∆x (or d) Consider a random, uniform accl object. Use the area of a trapezoid The second new formula

  8. Example #3 • A plane is moving at a speed of 500m/hr when it lands on a runway. Accelerating uniformly, it comes to a stop after covering 15m. How long did it take to stop? • Vi = 500 m/s • Vf = 0 • t = ? • a = ? • ∆x = 15 m ∆x =½∆t(vf+vi) 15 m = ½(t)(0+500 m/hr) 30 m = 500m/hr(t) t = 0.06 hr = 216 sec

  9. Example #3 • A plane is moving at a speed of 500m/hr when it lands on a runway. Accelerating uniformly, it comes to a stop after covering 15m. How long did it take to stop? What was its acceleration? • Vi = 500 m/hr • Vf = 0 • t = 0.03 hr or 108 s (previous slide) • a = ? • ∆x = 15 m • a = vf-vi t = 0-500m/hr 0.06 hr =-8333.33 m/hr2 Or -0.0006 m/s2

  10. HW • In book, • Complete pg 53,55, & 58 odds

  11. Day #2 • RE-Introduction to the equation of motion

  12. …. area under curve =

  13. 2. The first answer doesn’t make sense because you can’t have a negative time. You have to use the quadratic formula to solve the problem.

  14. a = vf-vi (rearrange) t Substitute our newest formula into the derivation below

  15. Make a decision as to which direction is POSITIVE & which is NEGATIVE List what you have (using “COMPATIBLE” units & using proper SIGNS) choose your equation Solve the equation Make sure your answer makes sense (both in MAGNITUDE & DIRECTION)

  16. 3.

  17. 4.

  18. Review – Solve • Solve the following: • You roll a ball up an incline at a speed of 4.5 m/s. After 5 sec, the ball is on the way down travelling at a new speed of 1.5 m/s. Find the ball’s acceleration. (Sketch) • -1.5-4.5 -6 -1.2 m/s2 5 5 rookie mistake…. 1.5-4.5 • Find the time to the peak. • -1.2 = 0-4.5 -1.2 = -4.5 =3.75 sec t t How far up does it go? 0 = (4.5)2 + 2(-1.2)d =8.44 m

  19. Day #3 • Free Fall

  20. Free Fall • Free fall problems are acceleration problems where we know the acceleration. • For all free fall problems (at the surface of the earth), • a = 9.8 m/s2 (down) the object speeds up by 9.8m/s2 after every second that it falls • When solving: • Make a chart • Use arrows next to variables to show the direction of the vectors • Fill in known variables • Solve for unknowns

  21. Free Fall Problems • Initial Velocity (Vi) will be 0, so our formulas change slightly. No need to memorize these.. • a = vf-vi t • a = vf t

  22. 9.8 m/s2 0 m/s 9.8 m/s 19.6 m/s 29.4 m/s 39.2 m/s 49 m/s

  23. Free Fall Problems • 1. A ball is dropped from the top of a bridge, and it takes 8 seconds to hit the water. • A. How tall was the bridge? • B. How fast was the ball going just before it hits the water? a. height? So find ∆x. b. a = (vf-vi)/t ∆x = vi ∆t +1/2(a)(∆t )2 9.8= vf/8 = ½(9.8)(8) 2 = 78.4 m/s = 313.6 m downward *You could solve part B first and then use the equation ∆x=1/2t(vi+vf)

  24. vi= 0 a = -9.8 m/s2 Dx = -20 m Because the diver is falling DOWNWARD, we can also use a negative sign

  25. If all objects accelerate at the same rate when falling near the earth’s surface, why do some objects actually hit the ground faster than others? (assuming they are dropped at the same time from the same height)? • AIR RESISTANCE!! • Some objects are narrower (like pencils) and therefore the air resists their falling less. • Some things are wider (like frisbees) and the air provides more resistance and slows down their fall. • NOTICE HOW THE MASS OF THE OBJECTS DOES NOT MATTER!

  26. 5. Sometimes the variable “y” is used instead of “x” to simply show that the object is moving vertically y v t t Notice that the object speeds up Notice that the acceleration is CONSTANT and NEGATIVE the ENTIRE way

  27. Day #4 • Throw-ups, Come-downs, Throw-downs

  28. Free Fall Problem – Throw up • “A ball is thrown up from the ground…” • In this type of problem, the initial velocity is always up. Also, the primary type of problem will ask “what is the maximum height”, which means vf = 0 m/s. • 2. A ball is thrown up from the ground at an initial speed of 20 m/s. • A. what is the maximum height the ball reaches? • B. How long before it gets to the max altitude? A. ∆x =? B. ∆t =? Vf2=vi2 +2a ∆x a=(vf-vi)/ ∆ t 0=2o2 +2(-9.8) ∆x -9.8=(0-20) ∆ t ∆x = 20.4m ∆ t=2.04s *You can solve for B first then use the formula ∆x =1/2t(vi+vf)

  29. Problem #2 2 UP +, DOWN - a = -9.8 m/s2 v2 = 0 (at the top) a) vi= 20 m/s 1 b) a = -9.8 m/s2 vf= 0 m/s c) To solve for “t”, use the quadratic formula  t= 0.66s and 3.42 sec (going up) (going down)

  30. Throw –up/Come down 3 scenarios 1. 2. 3. • Assume: • Up+, Down – • A =-9.8m/s2 • v2= 0 (at the top) • We’ll start with scenario #3. V3=-V1

  31. Scenario 3 – object is thrown up and caught at the same height 2 b) vf= 0 m/s a) 3 a = -9.8 m/s2 vf= 0 m/s 1 a = -9.8 m/s2 vi= 24.25 m/s Dx = 30 m c) Due to the fact that the ball is thrown and caught at the same height …

  32. Scenario 2 and part d of our problem 2 Dx = -1 m (because down is -) vi= 24.25 m/s a = -9.8 m/s2 vf2=vi2 +2a∆x vf2=24.252 +2(-9.8)(-1) 24.65 m/s down Or-24.65m/s 3 1 4

  33. What if we were solving for the time it was when the object hits the ground? vi= 24.25 m/s a = -9.8 m/s2 Dx = -1 m (because down is -) To solve for “t”, use the quadratic formula  Only positive times make sense

  34. Day #5 • Chase Problems

  35. Chase Problems • A chase problem is a scenario where two objects are involved, and they have the same position at some later time. • They can • Both start at the same place and same time • Start at different places but at the same time • Both start at the same place but at different times • Start at different places at different times • B or D but be moving in different directions

  36. The Key Equation • ∆x=1/2at2+vit ∆x=xf-xi • Substitute the second formula into the first, creating: x2=1/2at2+vit+xi • This formula will give you the final position for both objects. Since you don’t know that final position, set the two formulas equal to each other.

  37. Chase Problem #1 • Timmy is running at a constant speed of 6m/s. He sees Susie running 50m in front of him. She is moving at a constant speed of 4 m/s, in the same direction. How long will it take for Timmy to catch Susie? • Notice that while Timmy and Susie have different starting positions (xiT=0m while xiS=50m), they have the same final position (xf) after Timmy catches Susie. . .X xiTxiS x2 Tim = Susie 1/2at2+vit+xi = 1/2at2+vit+xi ½(0)t2+6t+0 =1/2(0)t2+4t=50 6t = 4t=50 t = 25sec

  38. Chase Problem #2 • A continuation of the last problem. • Timmy passes Susie, running at his constant speed of 6m/s. Susie decides to pick up the pace, very gradually. She begins to accelerate at a constant rate of 0.1m/s2. How long will it take her to catch Timmy? • Notice that Timmy and Susie have the same starting position (xiT=0m xiS=0m) and ending position (xf) after Susie catches Timmy. .x xit=xis x2 Tim = Susie 1/2at2+vit+xi = 1/2at2+vit+xi ½(0)t2+6t+0 = 1/2(0.1)t2+4t+0 6t=0.05t2+4t 6=0.05t+4 t=40sec Really, you could have made the initial position anything you wanted – they cancel out. I decided to choose 0.

  39. Chase Problem #3 + - 0 • A man drops a penny off the top of a 100m tall building. Exactly 1 second later another man throws a nickel downward from the same place as the first man. What is the minimum speed the nickel must be thrown at in order to catch the penny? • Start with the time it takes the penny to hit the bottom using the formula: ∆x=1/2at2+vit -100=1/2(-9.8)t2+0t t= 4.52 sec penny = nickel (1 sec later) 1/2at2+vit+xi = 1/2at2+vit+xi ½(-9.8)(4.52)2+0(4.52)+0 =1/2(-9.8)(3.52)2+vi(3.52)+0 Vi= -11.19 m/s (neg. because it’s going down)

  40. Chase Problem #4 really hard • In a strange, yet exciting, crash-test-dummy crash, two cars start by facing each other 1000m apart on a straight road. The first car accelerates from rest with a constant accel of 4m/s2. The second car accel, at a rate of 8 m/s2 for 5 sec but then settles into a constant speed. Find the elapsed time before these two cars collide. • Notice that the dummies have different starting positions (xiA=0, xiB=1000m), but they have the same final position (x2) when they crash. • Start by finding the final velocity of car B by using the following formula: • a = vf-vi-8=(vf-0)/5 so vf=-40m/s t 1/2at2+vit+xi = 1/2at2+vit+xi

  41. Test Review

  42. Step #1 Write Down What You Have (Look for “Key” Words)

  43. vf= 0 “Coming to a stop”“Starting from rest”“Coasting”“Maximum Height”“Dropped” vi= 0 vi= vf= constant vf= 0 a = -9.8 m/s2

  44. “Slowing Down”“Braking”“Speeding up”“Accelerating from rest” a = - __ a = + __

  45. Step #3 Solve the Equation Step #4 Make sure your answer makes sense

  46. Helpful Tip #1 Choose your “Key Points” in every problem…and do so wisely.

  47. Vertical Problems “Free-fall”“Throw up” “Throw-downs” 1 vi= 0 1 vi0 2 2 2 vf= 0 (at top) 1

  48. Vertical Problems “Throw up / Come Down” (throw and catch at same height) 2 vf= 0 (at top) 3 v1 =-v3 1

  49. “Throw up / Come Down” (throw and catch at different heights) Use and solve quadratically for “t” 2 2 Dx13 = + 3 1 Dx13 = - 1 3

  50. Helpful Tip #2 Assign positive and negative to different directions.

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