Phys 1501Q 1st Edition Lecture 2Outline of Last Lecture: Intro LectureI. Why study physics?II. Skills gained from studying physicsIII. Measurement UnitsIV. SI Unit PrefixesV. Basic Conversion OutlineOutline of Current Lecture: KinematicsI. KinematicsII. Position III. DisplacementIV. Velocitya. Average Velocityb. Instantaneous Velocityc. Constant Velocityd. Integrate VelocityV. Accelerationa. Average Accelerationb. Instantaneous Accelerationc. Constant Accelerationd. Integrate AccelerationVI. Velocity As A Function of TimeVII. Find Position as a Function of Timea. Checkpoint Problem: Ball on RampVIII. Find Velocity as a Function of Positiona. Checkpoint Problem: Racecar AccelerationIX. Free Falla. Checkpoint Problem: Object Droppedb. Checkpoint Problem: Object Thrown Up and DroppedThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Current Lecture: KinematicsI. KinematicsThe study of motionII. Position - Where an object is located (x)- Measured relative to a fixed point- Units = m (length)III. Displacement- Change in positionΔX=Xfinal – XinitialIV. Velocity- How an object’s motion changes over time- Distance/time or length/time- Units = m/s- Velocity has either a positive or negative direction- Speed is the absolute value of velocity or magnitude |v|Velocity (pink) is the slope of the position line (black)a. Average VelocityTotal displacement/total time ΔX/ΔYSlope of position line between 2 pointsCan be written as Vavgb. Instantaneous VelocityVelocity at a specific timeThe slope of the position curve at 1 point (aka tangent line to the curve)Also known as the derivative (dx/dt)c. Constant VelocityΔX = v*Δtd. Integrate VelocityGet a position from a known velocityV. Accelerationa. Average AccelerationSlope of velocity line between 2 pointsb. Instantaneous AccelerationSlope of velocity at a certain pointc. Constant AccelerationThe acceleration stays the same value, like gravity (-9.81m/s)d. Integrate AccelerationTo find an unknown velocityVI. Velocity As A Function of TimeV = Vo + atVII. Find Position as a Function of TimeVavg = Vot+1/2at^2a. Checkpoint Problem: Ball on RampA boll is rolled down a ramp with a constantacceleration. Between t=0 and t=1 sec it moves 1foot. How much does it move between t=1 andt=2?X(t)=1/2at^2X(1)1/2a(1)^2A=2X(2)=1/2(2)(2)^2 = 4ft total movedX(2) – x(1) = 3ft moved between t=1 and t=2VIII. Find Velocity as a Function of PositionSomething is undergoing constant accelerationWe do not know timeWe want velocity as a function of xv^2 = Vo^2 +2a(X-Xo)(Final velocity squared = initial velocity squared + 2 times acceleration * final minus initial position)a. Checkpoint Problem: Racecar AccelerationA racecar starts from rest and goes 400 m in 5 seconds. How fast was it going?Find 1)The final velocity 2)The average velocityo Find accelerationX – Xo = Vot + ½at^2Xo = 0 Vo=0 Now you have X = ½at^2a = 2x/t^2a = 2(400)/5^2 a = 32 m/s^2o 2. Find final velocityV = Vo + atV = 0 +32*5V=160 m/so 3. Find average velocity (Vavg)Vavg = V+Vo / 2Average velocity = 160 – 0/2 80 m/s = VavgIX. Free FallConstant acceleration of gravity acting on an objecta. Checkpoint Problem: Object DroppedAn object is dropped from 50 m high.1) When is impact? (How long does it take to hit?)2) What is impact velocity? (How fast is it going when it hits the ground?We know displacement = 50m acceleration= -9.8m/s^2X-Xo = Vot + ½at^20 – 50 = 0*t + ½(-9.8)t^2o V=Vo + at1. Solve for tt = sqrt((2*-50)/-9.8) t=3.2 seco 2. Find velocityV = Vo + atV = 0 + (-9.8)(3.2)Velocity = -31 m/sb. Checkpoint Problem: Object Thrown Up THEN DroppedA watermelon is tossed up from 50 ft, rises 10 m and then starts to fall. Y – Yo = (v^2 –Vo^2)/2ao 1. Find velocity when tossed upV^2 – Vo^2 = 2a(Y – Yo)V = 0Vo = sqrt(-2*-9.8*10)Vo = 14 m/so 2. Find time it takes to get to street levelY-Yo = Vot + ½at^2 0 = Vot + ½at^2 – (Y- Yo)Solve like a quadratic equation (find roots)-b +/- sqrt(b^2 – 4ac) / 2at = -2.1 or 4.9 (we can’t have negative time so t= 4.9 seconds)o 3. Find final velocityV = Vo + atV = 14+ (-9.8)(4.9) final velocity = -34 m/sNote:Throwing a ball 10 ft up from 50 ft is the same final velocity as dropping from 60
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