Phy 2053 Announcements Homework (“intro to webassign”) due tomorrow at 11:59pm is practice and not for credit. You can access homeworks without having a registration code (until Jan 17th). Homework 1 due Jan 21 is posted in webassign. It will count towards course grade. Solution manual: We are working with UF bookstore to order them for you to purchase. Yellow book (past exams and solutions) available at Target Copy for $16 Clicker questions will count towards course grade starting Jan 27. Instantaneous Velocity: keep making ∆t smaller: ∆x ∆t The slope of the line tangent to the position-vs.time graph is defined to be the instantaneous velocity at that time Acceleration ∆t VAB = ∆XAB/∆tAB = (Xf-Xi)/∆t = (54 m - 30 m)/10s = 2.2 ms-1 Typical Homework Problem 2.19 Runner A is initially 4.0 mi west of a flagpole and is running with a constant velocity of 6.0 mi/h due east. Runner B is initially 3.0 mi east of the flagpole and is running with a constant velocity of 5.0 mi/h due west. How far are the runners from the flagpole when they meet? Let’s pick it apart Average Acceleration Changing velocity means an acceleration is present Acceleration is the rate of change of the velocity a= Average vs. Instantaneous Velocity Average velocity between A and B v ∆x ∆x v vaverage = ∆t ∆v v f − v i = ∆t tf − ti Units are m/s² (SI), cm/s² (cgs), and ft/s² (US Cust) Vector quantity 1 Instantaneous acceleration = slope of tangent of velocity-time graph Position (m) velocity (m/s) acceleration (m/s2) Acceleration When the sign of the velocity and the acceleration are the same (either positive or negative), then the speed is increasing When the sign of the velocity and the acceleration are in the opposite directions, the speed is decreasing Kinematic Equations Used in situations with uniform acceleration v = v o + at ∆x = v o t + A negative acceleration does not necessarily mean the object is slowing down If the acceleration and velocity are both negative, the object is speeding up (1) 1 2 at 2 v 2 = v o2 + 2 a∆ x (2) (3) ⎛ v + vf ⎞ ∆x = v average t = ⎜ o ⎟t ⎝ 2 ⎠ Important: use the correct sign for x, v and a. Some examples of Use ∆x = v average Acceleration not in equation ⎛ v + vf ⎞ t=⎜ o ⎟t ⎝ 2 ⎠ Free Fall Displacement not in equation: v = vo + at Final velocity not in equation : ∆x = v o t + 1 2 at 2 If only force on object moving near surface of earth is gravity it is in free fall Free fall is constant acceleration The acceleration is called the acceleration due to gravity, symbolized by g g = 9.80 m/s² g is always directed downward Time not in equation : v 2 = v o2 + 2a∆x When estimating, use g ≈ 10 m/s2 toward the center of the earth Ignore air resistance and assume g doesn’t vary with altitude over short vertical distances Forces can apply to the object before or after the free fall 2 Free Fall options • Initial velocity is zero Problem-Solving Hints v=0 • Throw up -initial velocity non-zero and positive— instantaneous velocity at maximum height = 0 • Throw down –initial velocity is negative Convert if necessary Choose the appropriate kinematic equation Solve for the unknowns Choose a coordinate system, label initial and final points, indicate a positive direction for velocities and accelerations Label all quantities, be sure all the units are consistent • Starting and ending heights may be equal – symmetric or not equal-asymmetric - trajectory Read the problem Draw a diagram You may have to solve two equations for two unknowns Check your results Estimate and compare Check units A Typical Problem 2-53. A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engines stop at an altitude of 150 m. (a) what can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after lift-off does the rocket reach its maximum height? (d) How long is the rocket in the air? 3