PHYS 1441 – Section 002 Lecture #5AnnouncementsReminder: Special Problems for Extra CreditKinematic Equations of Motion on a Straight Line Under Constant AccelerationHow do we solve a problem using the kinematic formula for constant acceleration?Example 2.8Falling MotionExample for Using 1D Kinematic Equations on a Falling objectExample of a Falling Object cnt’d2D Coordinate SystemsExampleVector and ScalarProperties of VectorsVector OperationsExample for Vector AdditionComponents and Unit VectorsMonday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu1PHYS 1441 – Section 002Lecture #5Monday, Sept. 20, 2010Dr. Jaehoon Yu•One Dimensional Motion•One dimensional Kinematic Equations•How do we solve kinematic problems?•Falling motions•Vectors and Scalar•Properties and operations of vectors•Understanding a 2 Dimensional Motion•2D Kinematic Equations of MotionToday’s homework is homework #3, due 10pm, Tuesday, Sept. 28!!Announcements•1st term exam–Non-comprehensive –Time: 1 – 2:20pm, Wednesday, Sept. 22–Coverage: Appendices A.1 – A.8 and CH1.1 – CH3.4Monday, Sept. 20, 2010 2PHYS 1441-002, Fall 2010 Dr. Jaehoon YuMonday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu3Reminder: Special Problems for Extra Credit•Derive the quadratic equation for yx2-zx+v=0 5 points•Derive the kinematic equation from first principles and the known kinematic equations 10 points•You must show your OWN work in detail to obtain the full credit•Must be in much more detail than in pg. 19 of last lecture note!!!•Due by class Monday, Sept. 27 v2=v02+2a x−x0( )Monday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu4Kinematic Equations of Motion on a Straight Line Under Constant Acceleration tavtvxxixf221tatvxxxxiif xf−xi=vxt =12vxf+vxi( ) t ifxfxxavvxxi 222Velocity as a function of timeDisplacement as a function of velocities and timeDisplacement as a function of time, velocity, and accelerationVelocity as a function of Displacement and accelerationYou may use different forms of Kinetic equations, depending on the information given to you for specific physical problems!!Monday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu5How do we solve a problem using the kinematic formula for constant acceleration?•Identify what information is given in the problem.–Initial and final velocity?–Acceleration?–Distance?–Time?•Identify what the problem wants you to find out.•Identify which kinematic formula is most appropriate and easiest to solve for what the problem wants.–Often multiple formulae can give you the answer for the quantity you are looking for. Do not just use any formula but use the one that make the problem easiest to solve.•Solve the equation for the quantity wantedMonday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu6Example 2.82xfv = xf −xi =xiv =xfv =Suppose you want to design an air-bag system that can protect the driver in a head-on collision at a speed 100km/hr (~60miles/hr). Estimate how fast the air-bag must inflate to effectively protect the driver. Assume the car crumples upon impact over a distance of about 1m. How does the use of a seat belt help the driver? t =How long does it take for the car to come to a full stop? As long as it takes for it to crumple. We also know thatandUsing the kinematic formulaThe acceleration isxa =Thus the time for air-bag to deploy isThe initial speed of the car is 0m / s 100km / h = 100000m3600s=28m/ s1m vxi2+2axxf−xi( ) vxf2−vxi22 xf−xi( )= 0 − 28m/ s( )22×1m= −390m/ s2 vxf−vxia= 0 −28m/ s−390m/ s2= 0.07sMonday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu7Falling Motion•Falling motion is a motion under the influence of the gravitational pull (gravity) only; Which direction is a freely falling object moving?–A motion under constant acceleration –All kinematic formula we learned can be used to solve for falling motions. •Gravitational acceleration is inversely proportional to the distance between the object and the center of the earth•The magnitude of the gravitational acceleration is g=9.80m/s2 on the surface of the earth, most of the time.•The direction of gravitational acceleration is ALWAYS toward the center of the earth, which we normally call (-y); where up and down direction are indicated as the variable “y”•Thus the correct denotation of gravitational acceleration on the surface of the earth is g=-9.80m/s2 when +y points upwardYes, down to the center of the earth!!Monday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu8Example for Using 1D Kinematic Equations on a Falling object A stone was thrown straight upward at t=0 with +20.0m/s initial velocity on the roof of a 50.0m high building, g=-9.80m/s2fv =fy =(a) Find the time the stone reaches at the maximum height.What is the acceleration in this motion?What happens at the maximum height? The stone stops; V=0t =(b) Find the maximum height. vyi +ayt = +20.0−9.80t = 0.00m / sSolve for t 20.09.80= 2.04s 50.0 +20×2.04+12×(−9.80)×(2.04)2 =50.0+20.4=70.4(m) yi+vyit +12ayt2=Monday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu9Example of a Falling Object cnt’dt =yfv =fy =Positionyfv =Velocity(c) Find the time the stone reaches back to its original height.(d) Find the velocity of the stone when it reaches its original height.(e) Find the velocity and position of the stone at t=5.00s. 2.04 ×2=4.08s vyi +ayt = 20.0 +(−9.80)×4.08= −20.0(m/ s) vyi+ayt = 20.0 +(−9.80)×5.00=−29.0(m/ s) =50.0+20.0×5.00+12×(−9.80)×(5.00)2 yi+vyit +12ayt2 =+27.5(m)Monday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu102D Coordinate Systems•They make it easy and consistent to express locations or positions•Two commonly used systems, depending on convenience, are–Cartesian (Rectangular) Coordinate System•Coordinates are expressed in (x,y)–Polar Coordinate System •Coordinates are expressed in distance from the origin ® and the angle measured from the x-axis, θ(rθ)•Vectors become a lot easier to express and computeO (0,0)(x1,y1)r1θ11x =How are Cartesian and Polar coordinates related?1r =y1x1+x+y1y = tanθ1= x12+y12( )1y r1 r1=(r1θ) 1x θ1=tan−1y1x1⎛⎝⎜⎞⎠⎟ cosθ1 sinθ1Monday, Sept. 20, 2010 PHYS 1441-002, Fall 2010 Dr. Jaehoon Yu11Example Cartesian Coordinate of a point in the xy plane are (x,y)=
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