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# Clickers Setup

Clickers Setup. Turn on your clicker (press the power button) Set the frequency: Press and hold the power button Two letters will be flashing If it’s not “AC”, press “A” and then “C” If everything works, you should see “Welcome” and “Ready”. Kinematics of 1-dimensional motion. Télécharger la présentation ## Clickers Setup

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### Presentation Transcript

1. Clickers Setup • Turn on your clicker (press the power button) • Set the frequency: • Press and hold the power button • Two letters will be flashing • If it’s not “AC”, press “A” and then “C” • If everything works, you should see “Welcome” and “Ready”

2. Kinematics of 1-dimensional motion Displacement: Average velocity: Instantaneous velocity: Average acceleration: Instantaneous acceleration: Curvature

3. Equations of Kinematics for Constant Acceleration Parabola Proof: Result: Position x, velocity vx, acceleration ax, and time t are related by the kinematic equations. There are two initial conditions x0 and v0x .

4. Freely Falling Bodies Acceleration due to gravity g=9.8 m/s2 Example 2.6: Find velocity and position of a coin freely falling from rest after 1, 2, 3 s. • Solution: • Choose axes and plot trajectory. • Indicate known and unknown data By choice of axis Implied data 3. Write and solve equations in symbols 4. Do numerical calculations. 5. Check the answer: dimensions, signs, functional dependencies, scales…

5. Problem Solving Steps 1. Geometry & drawing:trajectory, vectors, coordinate axes free-body diagram, … 2. Data:a table of known and unknown quantities, including “implied data”. 3. Equations ( with reasoning comments ! ), their solution in algebraic form, and the final answers in algebraic form !!! 4. Numerical calculations and answers. 5. Check: dimensional, functional, scale, sign, … analysis of the answers and solution.

6. Exam Example 1 : Coin Toss Vy=0 y (problem 2.85) Questions: (a) How high does the coin go? 0 (b) What is the total time the coin is in the air? Total time T= 2 t = 1.2 s (c) What is its velocity when it comes back at y=0 ? for y=0 and vy<0 yields vy2 = v02→ vy = -v0 = - 6m/s

7. Position and Velocity by Integration Constant-acceleration formulas via integration

8. Example of kinematic problem with two possible solutions Accelerating spacecraft data Question: Find velocity at a displacement x=215 km ? x X=215 km = 215 000 m

9. Two segments of motion with different accelerations Segment 1 Segment 2 Question: Find total displacement D ? 0 x x1=x02 Segment 1 Segment 2

10. Exam Example 2: Accelerated Car (problems 2.7 and 2.17) Data: x(t)=αt+βt2+γt3, α=6m/s, β=1m/s2, γ = -2 m/s3, t=1s Find: (a) average and instantaneous velocities; (b) average and instantaneous accelerations; (c) a moment of time ts when the car stops. Solution: (a) v(t)=dx/dt= α+2βt +3γt2 ; v0=α; (b) a(t)=dv/dt= 2β +6γt; a0=2β; (c) v(ts)=0 → α+2βts +3γts2=0 0 x V(t) α 2β 0 t ts a(t)

11. Exam Example 3: Truck vs. Car (problem 2.34) Data:Truck v=+20 m/s Car v0=0, ac=+3.2 m/s2 0 Questions: (a) x where car overtakes the truck; (b) velocity of the car Vc at that x; (c) x(t) graphs for both vehicles; (d) v(t) graphs for both vehicles. x Solution: truck’s position x=vt,car’s positionxc=act2/2 • x=xc when vt=act2/2 → t=2v/ac → x=2v2/ac • (b) vc=v0+act → vc=2v x V(t) truck vc=2v car car truck v 0 t=2v/ac t 0 t v/ac t=2v/ac

12. Exam Example 4: Free fall past window (problem 2.84) Data: Δt=0.42 s ↔ h=y1-y2=1.9 m, v0y=0, ay= - g y Find: (a) y1 ; (b) v1y ; (c) v2y 0 V0y=0 1st solution: (b) Eq.(3) y2=y1+v1yΔt – gΔt2/2 → v1y= -h/Δt + gΔt/2 (a)Eq.(4) → v1y2= -2gy1 → y1 = - v1y2 /2g = -h2/[2g(Δt)]2 +h/2 – g(Δt)2 /8 (c) Eq.(4) v2y2 = v1y2 +2gh = (h/Δt + gΔt/2)2 y1 ay V1y h y2 V2y 2nd solution: • Free fall time from Eq.(3): t1=(2|y1|/g)1/2 , t2=(2|y2|/g)1/2→ Δt+t1=t2 (b) Eq.(4)→ (c) Eq.(4) →

13. Exam Example 5: Relative motion of free falling balls (problem 2.94) y 2 H Data: v0=1 m/s, H= 10 m, ay= - g Find: (a) Time of collision t; (b) Position of collision y; (c) What should be H in order v1(t)=0. 0 1 Solution: (a) Relative velocity of the balls is v0 for they have the same acceleration ay= –g → t = H/v0 (b) Eq.(3) for 2nd ball yields y = H – (1/2)gt2 = H – gH2/(2v02) (c) Eq.(1) for 1st ball yields v1 = v0 – gt = v0 – gH/v0 , hence, for v1=0 we find H = v02/g

14. Pulsars – Rotating Neutron Stars Discovery (1967) : “Little green men” Radio, optical, and X-ray pulses t T Period T~ 1ms– 4s is astonishingly stable !!! Pulse duration Neutron star Photons n, p+,e- Earth Compare: Rsun~ 700 000 km 10 km

15. Pulsar maps have been included on the two Pioneer Plaques as well as the Voyager Golden Record. They show the position of the Sun, relative to 14 pulsars, so that our position both in space and in time can be calculated by potential extraterrestrial intelligences. Pulsar positioning could create a spacecraft navigation system independently, or be an auxiliary device to GPS instruments.

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