DOC PREVIEW
UConn PHYS 1501Q - Kinematics
Type Lecture Note
Pages 6

This preview shows page 1-2 out of 6 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 6 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

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 xv^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


View Full Document

UConn PHYS 1501Q - Kinematics

Type: Lecture Note
Pages: 6
Download Kinematics
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Kinematics and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Kinematics 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?